Structure of atom-Exercises

Q1. (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.

Solution

Solution:

Given data:

Mass of one electron, \( m_e = 9.109 \times 10^{-28} \text{ g} \)
Charge of one electron, \( e = -1.602 \times 10^{-19} \text{ C} \)
Avogadro constant, \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \)

(i) Number of electrons whose total mass is 1 gram

Let the number of electrons be \( n \). Then,

\[ \begin{aligned} n \times m_e &= 1 \text{ g} \\ n &= \frac{1}{m_e} \\ n &= \frac{1}{9.109 \times 10^{-28}} \\ n &= 1.097 \times 10^{27} \end{aligned} \]

Thus, approximately \( 1.10 \times 10^{27} \) electrons together weigh one gram.

(ii) Mass and charge of one mole of electrons

Mass of one mole of electrons:

\[ \begin{aligned} \text{Mass} &= m_e \times N_A \\ &= (9.109 \times 10^{-28}) \times (6.022 \times 10^{23}) \\ &= 5.486 \times 10^{-4} \text{ g} \end{aligned} \]

Therefore, the mass of one mole of electrons is \( 5.49 \times 10^{-4} \text{ g} \).

Charge of one mole of electrons:

\[ \begin{aligned} \text{Charge} &= e \times N_A \\ &= (-1.602 \times 10^{-19}) \times (6.022 \times 10^{23}) \\ &= -9.65 \times 10^{4} \text{ C} \end{aligned} \]

Hence, one mole of electrons has a mass of \( 5.49 \times 10^{-4} \text{ g} \) and a total charge of \( -9.65 \times 10^{4} \text{ C} \).

Q2. (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron = \(\mathrm{1.675 \times 10^{–27}\ kg})\).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP. Will the answer change if the temperature and pressure are changed ?

Solution

Solution:

(i) Total number of electrons in one mole of methane, \( \mathrm{CH_4} \)

One carbon atom contains 6 electrons and each hydrogen atom contains 1 electron. Hence, one molecule of methane contains

\[ \begin{aligned} \text{Electrons per molecule} &= 6 + 4(1) \\ &= 10 \end{aligned} \]

Since one mole contains \( N_A = 6.022 \times 10^{23} \) molecules, the total number of electrons is

\[ \begin{aligned} \text{Total electrons} &= 10 \times 6.022 \times 10^{23} \\ &= 6.022 \times 10^{24} \end{aligned} \]

Thus, one mole of methane contains \( 6.022 \times 10^{24} \) electrons.

(ii) Total number and total mass of neutrons in 7 mg of \( ^{14}\mathrm{C} \)

Carbon–14 has atomic number 6 and mass number 14, therefore neutrons per atom are

\[ \begin{aligned} \text{Neutrons per atom} &= 14 - 6 = 8 \end{aligned} \]

Given mass of sample \( = 7 \text{ mg} = 7 \times 10^{-3} \text{ g} \). Number of moles of \( ^{14}\mathrm{C} \) is

\[ \begin{aligned} \text{Moles} &= \frac{7 \times 10^{-3}}{14} \\ &= 5.0 \times 10^{-4} \end{aligned} \]

Number of atoms present:

\[ \begin{aligned} \text{Atoms} &= 5.0 \times 10^{-4} \times 6.022 \times 10^{23} \\ &= 3.011 \times 10^{20} \end{aligned} \]

Hence, total number of neutrons:

\[ \begin{aligned} \text{Neutrons} &= 8 \times 3.011 \times 10^{20} \\ &= 2.41 \times 10^{21} \end{aligned} \]

Mass of one neutron \( = 1.675 \times 10^{-27} \text{ kg} \). Therefore, total mass of neutrons is

\[ \begin{aligned} \text{Total mass} &= 2.41 \times 10^{21} \times 1.675 \times 10^{-27} \\ &= 4.04 \times 10^{-6} \text{ kg} \end{aligned} \]

So, 7 mg of \( ^{14}\mathrm{C} \) contains \( 2.41 \times 10^{21} \) neutrons having a combined mass of \( 4.04 \times 10^{-6} \) kg.

(iii) Total number and total mass of protons in 34 mg of \( \mathrm{NH_3} \)

In one molecule of ammonia, nitrogen contributes 7 protons and three hydrogen atoms contribute 3 protons. Hence, protons per molecule are

\[ \begin{aligned} 7 + 3 = 10 \end{aligned} \]

Molar mass of \( \mathrm{NH_3} = 17 \text{ g mol}^{-1} \). Given mass \( = 34 \text{ mg} = 0.034 \text{ g} \). Number of moles:

\[ \begin{aligned} \text{Moles} &= \frac{0.034}{17} \\ &= 2.0 \times 10^{-3} \end{aligned} \]

Number of molecules:

\[ \begin{aligned} \text{Molecules} &= 2.0 \times 10^{-3} \times 6.022 \times 10^{23} \\ &= 1.204 \times 10^{21} \end{aligned} \]

Therefore, total number of protons:

\[ \begin{aligned} \text{Protons} &= 10 \times 1.204 \times 10^{21} \\ &= 1.204 \times 10^{22} \end{aligned} \]

Taking mass of one proton as \( 1.673 \times 10^{-27} \text{ kg} \), total mass of protons is

\[ \begin{aligned} \text{Total mass} &= 1.204 \times 10^{22} \times 1.673 \times 10^{-27} \\ &= 2.01 \times 10^{-5} \text{ kg} \end{aligned} \]

Hence, 34 mg of ammonia contains \( 1.204 \times 10^{22} \) protons with a combined mass of \( 2.01 \times 10^{-5} \) kg.

The answers will not change with temperature or pressure, because the number of particles depends only on the given mass of the substance, not on external conditions.

Q3. How many neutrons and protons are there in the following nuclei ?
\(\mathrm{^{13}_6C,\; ^{16}_8O,\; ^{24}_{12}Mg,\; ^{56}_{26}Fe,\; ^{88}_{38}Sr}\)

Solution

Solution:

For any nucleus represented as \( ^A_ZX \), the atomic number \( Z \) gives the number of protons, while the number of neutrons is obtained from \( A - Z \), where \( A \) is the mass number.

For \( ^{13}_{6}\mathrm{C} \):

\[ \begin{aligned} \text{Protons} &= Z = 6 \\ \text{Neutrons} &= A - Z = 13 - 6 = 7 \end{aligned} \]

Thus, carbon-13 contains 6 protons and 7 neutrons.

For \( ^{16}_{8}\mathrm{O} \):

\[ \begin{aligned} \text{Protons} &= 8 \\ \text{Neutrons} &= 16 - 8 = 8 \end{aligned} \]

Therefore, oxygen-16 has 8 protons and 8 neutrons.

For \( ^{24}_{12}\mathrm{Mg} \):

\[ \begin{aligned} \text{Protons} &= 12 \\ \text{Neutrons} &= 24 - 12 = 12 \end{aligned} \]

Hence, magnesium-24 consists of 12 protons and 12 neutrons.

For \( ^{56}_{26}\mathrm{Fe} \):

\[ \begin{aligned} \text{Protons} &= 26 \\ \text{Neutrons} &= 56 - 26 = 30 \end{aligned} \]

So, iron-56 contains 26 protons and 30 neutrons.

For \( ^{88}_{38}\mathrm{Sr} \):

\[ \begin{aligned} \text{Protons} &= 38 \\ \text{Neutrons} &= 88 - 38 = 50 \end{aligned} \]

Accordingly, strontium-88 has 38 protons and 50 neutrons.

In summary, the nuclei contain the following particles: \( ^{13}_{6}\mathrm{C} \) has 6 protons and 7 neutrons, \( ^{16}_{8}\mathrm{O} \) has 8 protons and 8 neutrons, \( ^{24}_{12}\mathrm{Mg} \) has 12 protons and 12 neutrons, \( ^{56}_{26}\mathrm{Fe} \) has 26 protons and 30 neutrons, and \( ^{88}_{38}\mathrm{Sr} \) has 38 protons and 50 neutrons.

Q4. Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9

Solution

Solution:

The complete atomic symbol is written in the form \( ^A_ZX \), where \( A \) represents the mass number, \( Z \) represents the atomic number, and \( X \) is the chemical symbol of the element. The element is identified using its atomic number.

(i) For \( Z = 17 \) and \( A = 35 \):

Atomic number 17 corresponds to chlorine (Cl). Hence,

\[ \begin{aligned} \text{Complete symbol} &= ^{35}_{17}\mathrm{Cl} \end{aligned} \]

(ii) For \( Z = 92 \) and \( A = 233 \):

Atomic number 92 corresponds to uranium (U). Therefore,

\[ \begin{aligned} \text{Complete symbol} &= ^{233}_{92}\mathrm{U} \end{aligned} \]

(iii) For \( Z = 4 \) and \( A = 9 \):

Atomic number 4 corresponds to beryllium (Be). Thus,

\[ \begin{aligned} \text{Complete symbol} &= ^{9}_{4}\mathrm{Be} \end{aligned} \]

Hence, the required complete atomic symbols are \( ^{35}_{17}\mathrm{Cl} \), \( ^{233}_{92}\mathrm{U} \), and \( ^{9}_{4}\mathrm{Be} \).

Q5. Yellow light emitted from a sodium lamp has a wavelength \((\lambda)\) of 580 nm. Calculate the frequency \((\nu)\) and wavenumber \((\overline{\nu})\) of the yellow light.

Solution

Solution:

Given wavelength of yellow light, \( \lambda = 580 \text{ nm} = 580 \times 10^{-9} \text{ m} = 5.80 \times 10^{-7} \text{ m} \). The speed of light is \( c = 3.00 \times 10^{8} \text{ m s}^{-1} \).

First, the frequency \( \nu \) is calculated using the relation \( c = \lambda \nu \).

\[ \begin{aligned} \nu &= \frac{c}{\lambda} \\ &= \frac{3.00 \times 10^{8}}{5.80 \times 10^{-7}} \\ &= 5.17 \times 10^{14} \text{ s}^{-1} \end{aligned} \]

Thus, the frequency of the yellow light is \( 5.17 \times 10^{14} \text{ Hz} \).

Next, the wavenumber \( \overline{\nu} \) is defined as the reciprocal of wavelength expressed in centimetres. Converting wavelength into centimetres:

\[ \lambda = 5.80 \times 10^{-7} \text{ m} = 5.80 \times 10^{-5} \text{ cm} \]

Now,

\[ \begin{aligned} \overline{\nu} &= \frac{1}{\lambda} \\ &= \frac{1}{5.80 \times 10^{-5}} \\ &= 1.72 \times 10^{4} \text{ cm}^{-1} \end{aligned} \]

Hence, the yellow light from the sodium lamp has a frequency of \( 5.17 \times 10^{14} \text{ Hz} \) and a wavenumber of \( 1.72 \times 10^{4} \text{ cm}^{-1} \).

Q6. Find energy of each of the photons which
(i) correspond to light of frequency 3×1015 Hz.
(ii) have wavelength of 0.50 Å.

Solution

Solution:

The energy of a photon is given by Planck’s relation \( E = h\nu \), where \( h = 6.626 \times 10^{-34} \text{ J s} \) is Planck’s constant and \( \nu \) is the frequency of radiation.

(i) For light of frequency \( \nu = 3.0 \times 10^{15} \text{ Hz} \):

\[ \begin{aligned} E &= h\nu \\ &= (6.626 \times 10^{-34})(3.0 \times 10^{15}) \\ &= 1.99 \times 10^{-18} \text{ J} \end{aligned} \]

Hence, each photon corresponding to this frequency has an energy of \( 1.99 \times 10^{-18} \text{ J} \).

(ii) For radiation having wavelength \( \lambda = 0.50 \text{ Å} = 0.50 \times 10^{-10} \text{ m} = 5.0 \times 10^{-11} \text{ m} \), the frequency is first calculated using \( \nu = \frac{c}{\lambda} \), where \( c = 3.00 \times 10^{8} \text{ m s}^{-1} \).

\[ \begin{aligned} \nu &= \frac{3.00 \times 10^{8}}{5.0 \times 10^{-11}} \\ &= 6.0 \times 10^{18} \text{ Hz} \end{aligned} \]

Now substituting this value into Planck’s equation:

\[ \begin{aligned} E &= h\nu \\ &= (6.626 \times 10^{-34})(6.0 \times 10^{18}) \\ &= 3.98 \times 10^{-15} \text{ J} \end{aligned} \]

Therefore, the energy of each photon having wavelength \( 0.50 \text{ Å} \) is \( 3.98 \times 10^{-15} \text{ J} \).

Q7. Calculate the wavelength, frequency and wavenumber of a light wave whose period is \(\mathrm{2.0 \times 10^{–10}\ s}\).

Solution

Solution:

The given period of the light wave is \( T = 2.0 \times 10^{-10} \text{ s} \). The frequency is related to the period by the relation \( \nu = \frac{1}{T} \).

\[ \begin{aligned} \nu &= \frac{1}{T} \\ &= \frac{1}{2.0 \times 10^{-10}} \\ &= 5.0 \times 10^{9} \text{ Hz} \end{aligned} \]

Thus, the frequency of the light wave is \( 5.0 \times 10^{9} \text{ Hz} \).

The wavelength is calculated using the relation \( \lambda = \frac{c}{\nu} \), where the speed of light \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \).

\[ \begin{aligned} \lambda &= \frac{3.0 \times 10^{8}}{5.0 \times 10^{9}} \\ &= 6.0 \times 10^{-2} \text{ m} \end{aligned} \]

Hence, the wavelength of the light wave is \( 6.0 \times 10^{-2} \text{ m} \).

The wavenumber \( \overline{\nu} \) is the reciprocal of wavelength expressed in centimetres. First, convert wavelength into centimetres:

\[ \lambda = 6.0 \times 10^{-2} \text{ m} = 6.0 \text{ cm} \]

Now,

\[ \begin{aligned} \overline{\nu} &= \frac{1}{\lambda} \\ &= \frac{1}{6.0} \\ &= 1.67 \times 10^{-1} \text{ cm}^{-1} \end{aligned} \]

Therefore, the wavelength, frequency and wavenumber of the given light wave are \( 6.0 \times 10^{-2} \text{ m} \), \( 5.0 \times 10^{9} \text{ Hz} \), and \( 1.67 \times 10^{-1} \text{ cm}^{-1} \) respectively.

Q8. What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?

Solution

Solution:

The wavelength of the light is given as \( \lambda = 4000 \text{ pm} \). Since \( 1 \text{ pm} = 10^{-12} \text{ m} \), the wavelength in metres is

\[ \begin{aligned} \lambda &= 4000 \times 10^{-12} \\ &= 4.0 \times 10^{-9} \text{ m} \end{aligned} \]

The energy of one photon is calculated using the relation \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \text{ J s} \) and \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \).

\[ \begin{aligned} E &= \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{4.0 \times 10^{-9}} \\ &= \frac{1.9878 \times 10^{-25}}{4.0 \times 10^{-9}} \\ &= 4.97 \times 10^{-17} \text{ J} \end{aligned} \]

Thus, each photon of wavelength \( 4000 \text{ pm} \) has an energy of \( 4.97 \times 10^{-17} \text{ J} \).

If the total energy supplied is \( 1 \text{ J} \), the number of photons \( n \) required is

\[ \begin{aligned} n &= \frac{1}{E} \\ &= \frac{1}{4.97 \times 10^{-17}} \\ &= 2.01 \times 10^{16} \end{aligned} \]

Therefore, approximately \( 2.0 \times 10^{16} \) photons of wavelength \( 4000 \text{ pm} \) are required to provide 1 joule of energy.

Q9. A photon of wavelength \(\mathrm{4 \times 10^{–7}\ m}\) strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron \(\mathrm{(1\ eV= 1.6020 \times 10^{–19}\ J)}\).

Solution

Solution:

The wavelength of the incident photon is \( \lambda = 4.0 \times 10^{-7} \text{ m} \). Planck’s constant is \( h = 6.626 \times 10^{-34} \text{ J s} \) and the speed of light is \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \). The work function of the metal is given as \( \phi = 2.13 \text{ eV} \).

(i) Energy of the photon is calculated using \( E = \frac{hc}{\lambda} \):

\[ \begin{aligned} E &= \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{4.0 \times 10^{-7}} \\ &= 4.97 \times 10^{-19} \text{ J} \end{aligned} \]

Converting this energy into electron volts using \( 1 \text{ eV} = 1.6020 \times 10^{-19} \text{ J} \):

\[ \begin{aligned} E &= \frac{4.97 \times 10^{-19}}{1.6020 \times 10^{-19}} \\ &= 3.10 \text{ eV} \end{aligned} \]

Thus, the energy of the photon is approximately \( 3.10 \text{ eV} \).

(ii) According to Einstein’s photoelectric equation, the kinetic energy of the emitted electron is \( K.E. = E - \phi \).

\[ \begin{aligned} K.E. &= 3.10 - 2.13 \\ &= 0.97 \text{ eV} \end{aligned} \]

Therefore, the maximum kinetic energy of the photoelectron is \( 0.97 \text{ eV} \). In joules,

\[ K.E. = 0.97 \times 1.6020 \times 10^{-19} = 1.55 \times 10^{-19} \text{ J} \]

(iii) The velocity of the photoelectron is obtained from \( K.E. = \frac{1}{2}mv^2 \), where the mass of electron \( m = 9.11 \times 10^{-31} \text{ kg} \).

\[ \begin{aligned} v &= \sqrt{\frac{2K.E.}{m}} \\ &= \sqrt{\frac{2 \times 1.55 \times 10^{-19}}{9.11 \times 10^{-31}}} \\ &= \sqrt{3.40 \times 10^{11}} \\ &= 5.83 \times 10^{5} \text{ m s}^{-1} \end{aligned} \]

Hence, the photon has an energy of \( 3.10 \text{ eV} \), the emitted electron possesses a kinetic energy of \( 0.97 \text{ eV} \), and the velocity of the photoelectron is approximately \( 5.8 \times 10^{5} \text{ m s}^{-1} \).

Q10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in \(\mathrm{kJ\ mol^{–1}}\).

Solution

Solution:

The wavelength of the radiation is given as \( \lambda = 242 \text{ nm} = 242 \times 10^{-9} \text{ m} = 2.42 \times 10^{-7} \text{ m} \). Since this radiation is just sufficient to ionise sodium, the energy of one photon is equal to the ionisation energy of one atom. Planck’s constant is \( h = 6.626 \times 10^{-34} \text{ J s} \) and the speed of light is \( c = 3.00 \times 10^{8} \text{ m s}^{-1} \).

First, the energy of a single photon is calculated using \( E = \frac{hc}{\lambda} \).

\[ \begin{aligned} E &= \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{2.42 \times 10^{-7}} \\ &= \frac{1.9878 \times 10^{-25}}{2.42 \times 10^{-7}} \\ &= 8.22 \times 10^{-19} \text{ J} \end{aligned} \]

Thus, the ionisation energy per sodium atom is \( 8.22 \times 10^{-19} \text{ J} \). To obtain the ionisation energy per mole, this value is multiplied by Avogadro’s number, \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \).

\[ \begin{aligned} E_{\text{mol}} &= 8.22 \times 10^{-19} \times 6.022 \times 10^{23} \\ &= 4.95 \times 10^{5} \text{ J mol}^{-1} \end{aligned} \]

Converting joules into kilojoules,

\[ E_{\text{mol}} = 4.95 \times 10^{2} \text{ kJ mol}^{-1} \]

Hence, the ionisation energy of sodium is approximately \( 4.95 \times 10^{2} \text{ kJ mol}^{-1} \), or about \( 495 \text{ kJ mol}^{-1} \).

Q11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of emission of quanta per second.

Solution

Solution:

The power of the bulb is given as \( P = 25 \text{ W} \), which means the bulb emits energy at the rate of \( 25 \text{ J s}^{-1} \). The wavelength of the yellow light is \( \lambda = 0.57 \,\mu\text{m} = 0.57 \times 10^{-6} \text{ m} = 5.7 \times 10^{-7} \text{ m} \). Planck’s constant is \( h = 6.626 \times 10^{-34} \text{ J s} \) and the speed of light is \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \).

First, the energy of one quantum (photon) of this radiation is obtained from \( E = \frac{hc}{\lambda} \).

\[ \begin{aligned} E &= \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{5.7 \times 10^{-7}} \\ &= \frac{1.9878 \times 10^{-25}}{5.7 \times 10^{-7}} \\ &= 3.49 \times 10^{-19} \text{ J} \end{aligned} \]

Thus, each photon carries an energy of \( 3.49 \times 10^{-19} \text{ J} \).

Since the bulb emits \( 25 \text{ J} \) of energy every second, the number of photons emitted per second is

\[ \begin{aligned} N &= \frac{\text{energy per second}}{\text{energy per photon}} \\ &= \frac{25}{3.49 \times 10^{-19}} \\ &= 7.16 \times 10^{19} \end{aligned} \]

Therefore, the rate of emission of quanta is approximately \( 7.2 \times 10^{19} \) photons per second.

Q12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency \((ν_0)\) and work function \((W_0)\) of the metal.

Solution

Solution:

The given wavelength is \( \lambda = 6800 \,\text{Å} \). Since \( 1 \,\text{Å} = 10^{-10} \text{ m} \), we have

\[ \begin{aligned} \lambda &= 6800 \times 10^{-10} \\ &= 6.8 \times 10^{-7} \text{ m} \end{aligned} \]

Electrons are emitted with zero velocity, which means the incident radiation has just the threshold energy. Therefore, the frequency of the radiation equals the threshold frequency \( \nu_0 \). Using the relation \( \nu_0 = \frac{c}{\lambda} \), where \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \),

\[ \begin{aligned} \nu_0 &= \frac{3.0 \times 10^{8}}{6.8 \times 10^{-7}} \\ &= 4.41 \times 10^{14} \text{ Hz} \end{aligned} \]

Thus, the threshold frequency is \( 4.41 \times 10^{14} \text{ Hz} \).

The work function \( W_0 \) is given by \( W_0 = h\nu_0 \), where \( h = 6.626 \times 10^{-34} \text{ J s} \).

\[ \begin{aligned} W_0 &= (6.626 \times 10^{-34})(4.41 \times 10^{14}) \\ &= 2.92 \times 10^{-19} \text{ J} \end{aligned} \]

Therefore, the work function of the metal is \( 2.92 \times 10^{-19} \text{ J} \).

Q13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Solution

Solution:

For electronic transitions in hydrogen, the wavelength of emitted radiation is given by the Rydberg equation,

\[ \frac{1}{\lambda} = R\left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right) \]

where \( R = 1.097 \times 10^{7} \text{ m}^{-1} \), \( n_1 \) is the higher energy level, and \( n_2 \) is the lower energy level. Here, the electron falls from \( n_1 = 4 \) to \( n_2 = 2 \).

Substituting the values,

\[ \begin{aligned} \frac{1}{\lambda} &= 1.097 \times 10^{7}\left(\frac{1}{2^2} - \frac{1}{4^2}\right) \\ &= 1.097 \times 10^{7}\left(\frac{1}{4} - \frac{1}{16}\right) \\ &= 1.097 \times 10^{7}\left(\frac{3}{16}\right) \\ &= 2.06 \times 10^{6} \text{ m}^{-1} \end{aligned} \]

Now taking the reciprocal to obtain the wavelength,

\[ \begin{aligned} \lambda &= \frac{1}{2.06 \times 10^{6}} \\ &= 4.86 \times 10^{-7} \text{ m} \end{aligned} \]

Hence, the wavelength of light emitted during the transition from \( n = 4 \) to \( n = 2 \) is \( 4.86 \times 10^{-7} \text{ m} \), or equivalently \( 486 \text{ nm} \).

Q14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit).

Solution

Solution:

The energy of an electron in the \( n^{\text{th}} \) orbit of a hydrogen atom is given by \( E_n = -\dfrac{13.6}{n^2} \text{ eV} \). Ionisation corresponds to removing the electron completely, that is, raising its energy to zero.

First, consider ionisation from the \( n = 5 \) orbit:

\[ \begin{aligned} E_5 &= -\frac{13.6}{5^2} \\ &= -\frac{13.6}{25} \\ &= -0.544 \text{ eV} \end{aligned} \]

Since the final energy after ionisation is zero, the energy required to remove the electron from this orbit is

\[ \begin{aligned} \Delta E &= 0 - (-0.544) \\ &= 0.544 \text{ eV} \end{aligned} \]

Thus, only \( 0.544 \text{ eV} \) is needed to ionise hydrogen when the electron is in the \( n = 5 \) orbit.

Now, for comparison, the ionisation enthalpy of hydrogen from the ground state \( (n = 1) \) is

\[ \begin{aligned} E_1 &= -\frac{13.6}{1^2} \\ &= -13.6 \text{ eV} \end{aligned} \]

Therefore, the energy required to remove the electron from the \( n = 1 \) orbit is

\[ \Delta E = 13.6 \text{ eV} \]

Hence, ionisation from the \( n = 5 \) level requires only \( 0.544 \text{ eV} \), which is much smaller than the ground-state ionisation energy of \( 13.6 \text{ eV} \). This clearly shows that electrons in higher orbits are far less tightly bound to the nucleus than those in the lowest orbit.

Q15. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

Solution

Solution:

When an electron in a hydrogen atom is excited to a higher energy level, it can return to the ground state either directly or through several intermediate levels. Each possible transition between two energy levels produces one emission line.

If the electron is initially in the level \( n = 6 \), the total number of distinct emission lines corresponds to the total number of possible transitions among the six energy levels \( (1, 2, 3, 4, 5, 6) \). This is given by the relation

\[ \text{Number of lines} = \frac{n(n-1)}{2} \]

Substituting \( n = 6 \),

\[ \begin{aligned} \text{Number of lines} &= \frac{6(6-1)}{2} \\ &= \frac{6 \times 5}{2} \\ &= 15 \end{aligned} \]

Therefore, when an electron initially in the \( n = 6 \) level finally reaches the ground state, a maximum of 15 different emission lines can be observed.

Q16. (i) The energy associated with the first orbit in the hydrogen atom is \(\mathrm{–2.18 \times 10^{–18}\ J\ atom^{–1}}\). What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Solution

Solution:

(i) For a hydrogen atom, the energy of the electron in the \( n^{\text{th}} \) orbit is related to that of the first orbit by \( E_n = \dfrac{E_1}{n^2} \), where \( E_1 = -2.18 \times 10^{-18} \text{ J atom}^{-1} \).

For the fifth orbit \( (n = 5) \),

\[ \begin{aligned} E_5 &= \frac{-2.18 \times 10^{-18}}{5^2} \\ &= \frac{-2.18 \times 10^{-18}}{25} \\ &= -8.72 \times 10^{-20} \text{ J atom}^{-1} \end{aligned} \]

Hence, the energy associated with the fifth orbit of hydrogen is \( -8.72 \times 10^{-20} \text{ J atom}^{-1} \).

(ii) The radius of the \( n^{\text{th}} \) Bohr orbit is given by \( r_n = n^2 a_0 \), where \( a_0 = 5.29 \times 10^{-11} \text{ m} \) is the Bohr radius.

For \( n = 5 \),

\[ \begin{aligned} r_5 &= 5^2 \times 5.29 \times 10^{-11} \\ &= 25 \times 5.29 \times 10^{-11} \\ &= 1.32 \times 10^{-9} \text{ m} \end{aligned} \]

Therefore, the radius of Bohr’s fifth orbit for the hydrogen atom is \( 1.32 \times 10^{-9} \text{ m} \).

Q17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Solution

Solution:

In the Balmer series of hydrogen, all electronic transitions terminate at the level \( n_2 = 2 \). The longest wavelength corresponds to the smallest energy difference, which occurs for the transition from \( n_1 = 3 \) to \( n_2 = 2 \).

The wavenumber is given by the Rydberg relation,

\[ \overline{\nu} = R\left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right) \]

where \( R = 1.097 \times 10^{7} \text{ m}^{-1} \), \( n_1 = 3 \), and \( n_2 = 2 \). Substituting these values,

\[ \begin{aligned} \overline{\nu} &= 1.097 \times 10^{7}\left(\frac{1}{2^2} - \frac{1}{3^2}\right) \\ &= 1.097 \times 10^{7}\left(\frac{1}{4} - \frac{1}{9}\right) \\ &= 1.097 \times 10^{7}\left(\frac{5}{36}\right) \\ &= 1.52 \times 10^{6} \text{ m}^{-1} \end{aligned} \]

Hence, the wavenumber corresponding to the longest wavelength transition in the Balmer series is \( 1.52 \times 10^{6} \text{ m}^{-1} \).

Q18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is \(\mathrm{–2.18 \times 10^{–11}\ ergs}\).

Solution

Solution:

The energy of an electron in the \( n^{\text{th}} \) Bohr orbit of hydrogen is given by \( E_n = \dfrac{E_1}{n^2} \), where the ground state energy is \( E_1 = -2.18 \times 10^{-11} \text{ ergs} \). Since \( 1 \text{ erg} = 10^{-7} \text{ J} \), this becomes \( E_1 = -2.18 \times 10^{-18} \text{ J} \).

First, the energy of the fifth orbit is calculated.

\[ \begin{aligned} E_5 &= \frac{-2.18 \times 10^{-18}}{5^2} \\ &= \frac{-2.18 \times 10^{-18}}{25} \\ &= -8.72 \times 10^{-20} \text{ J} \end{aligned} \]

The energy required to raise the electron from the first orbit to the fifth orbit is the difference between these two energies:

\[ \begin{aligned} \Delta E &= E_5 - E_1 \\ &= (-8.72 \times 10^{-20}) - (-2.18 \times 10^{-18}) \\ &= 2.09 \times 10^{-18} \text{ J} \end{aligned} \]

Hence, \( 2.09 \times 10^{-18} \text{ J} \) of energy is required to excite the electron from \( n = 1 \) to \( n = 5 \).

When the electron returns directly from \( n = 5 \) to the ground state, the emitted photon carries the same energy \( \Delta E \). The wavelength of this radiation is obtained using \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \text{ J s} \) and \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \).

\[ \begin{aligned} \lambda &= \frac{hc}{E} \\ &= \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{2.09 \times 10^{-18}} \\ &= 9.52 \times 10^{-8} \text{ m} \end{aligned} \]

Therefore, the energy required is \( 2.09 \times 10^{-18} \text{ J} \), and the wavelength of light emitted when the electron falls back to the ground state is \( 9.52 \times 10^{-8} \text{ m} \), or about \( 95.2 \text{ nm} \).

Q19. The electron energy in hydrogen atom is given by \(\mathrm{E_n= (–2.18 \times 10^{–18})/n^2\ J}\). Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Solution

Solution:

The energy of an electron in the \( n^{\text{th}} \) orbit of hydrogen is given as \( E_n = \dfrac{-2.18 \times 10^{-18}}{n^2} \text{ J} \). For the second orbit \( (n = 2) \), the energy of the electron is

\[ \begin{aligned} E_2 &= \frac{-2.18 \times 10^{-18}}{2^2} \\ &= \frac{-2.18 \times 10^{-18}}{4} \\ &= -5.45 \times 10^{-19} \text{ J} \end{aligned} \]

To remove the electron completely from this orbit, its energy must be raised to zero. Hence, the energy required for ionisation from \( n = 2 \) is

\[ \begin{aligned} \Delta E &= 0 - (-5.45 \times 10^{-19}) \\ &= 5.45 \times 10^{-19} \text{ J} \end{aligned} \]

Therefore, \( 5.45 \times 10^{-19} \text{ J} \) of energy is required to completely remove the electron from the second orbit.

The longest wavelength of light capable of producing this transition corresponds to this minimum energy. Using the relation \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \text{ J s} \) and \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \),

\[ \begin{aligned} \lambda &= \frac{hc}{E} \\ &= \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{5.45 \times 10^{-19}} \\ &= 3.65 \times 10^{-7} \text{ m} \end{aligned} \]

Converting metres into centimetres,

\[ \lambda = 3.65 \times 10^{-7} \text{ m} = 3.65 \times 10^{-5} \text{ cm} \]

Hence, the energy required is \( 5.45 \times 10^{-19} \text{ J} \), and the longest wavelength of light that can cause this ionisation is \( 3.65 \times 10^{-5} \text{ cm} \).

Q20. Calculate the wavelength of an electron moving with a velocity of \(\mathrm{2.05 \times 10^7\ m\ s^{–1}}\).

Solution

Solution:

The de Broglie wavelength of a moving electron is given by \( \lambda = \dfrac{h}{mv} \), where \( h = 6.626 \times 10^{-34} \text{ J s} \) is Planck’s constant, \( m = 9.11 \times 10^{-31} \text{ kg} \) is the mass of an electron, and \( v = 2.05 \times 10^{7} \text{ m s}^{-1} \) is the velocity of the electron.

Substituting the given values,

\[ \begin{aligned} \lambda &= \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(2.05 \times 10^{7})} \\ &= \frac{6.626 \times 10^{-34}}{1.87 \times 10^{-23}} \\ &= 3.55 \times 10^{-11} \text{ m} \end{aligned} \]

Hence, the wavelength associated with the electron is \( 3.55 \times 10^{-11} \text{ m} \), or approximately \( 0.0355 \text{ nm} \).

Q21. The mass of an electron is \(\mathrm{9.1 \times 10^{–31}\ kg}\). If its K.E. is \(\mathrm{3.0 \times 10^{–25}\ J}\), calculate its wavelength.

Solution

Solution:

The de Broglie wavelength of a particle is given by \( \lambda = \dfrac{h}{mv} \). Since the kinetic energy is known, we first determine the velocity of the electron using \( K.E. = \dfrac{1}{2}mv^2 \).

Given \( m = 9.1 \times 10^{-31} \text{ kg} \) and \( K.E. = 3.0 \times 10^{-25} \text{ J} \),

\[ \begin{aligned} v &= \sqrt{\frac{2\,K.E.}{m}} \\ &= \sqrt{\frac{2 \times 3.0 \times 10^{-25}}{9.1 \times 10^{-31}}} \\ &= \sqrt{\frac{6.0 \times 10^{-25}}{9.1 \times 10^{-31}}} \\ &= \sqrt{6.59 \times 10^{5}} \\ &= 8.12 \times 10^{2} \text{ m s}^{-1} \end{aligned} \]

Now substituting this velocity into the de Broglie relation, with \( h = 6.626 \times 10^{-34} \text{ J s} \),

\[ \begin{aligned} \lambda &= \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31})(8.12 \times 10^{2})} \\ &= \frac{6.626 \times 10^{-34}}{7.39 \times 10^{-28}} \\ &= 8.97 \times 10^{-7} \text{ m} \end{aligned} \]

Therefore, the wavelength associated with the electron is \( 8.97 \times 10^{-7} \text{ m} \), which is approximately \( 8.97 \times 10^{-5} \text{ cm} \).

Q22. Which of the following are isoelectronic species i.e., those having the same number of electrons?
\(\mathrm{Na^+,\ K^+,\ Mg^{2+},\ Ca^{2+},\ S^{2–},\ Ar}\).

Solution

Solution:

Isoelectronic species are atoms or ions that possess the same total number of electrons. To identify such species, we first determine the number of electrons in each case using their atomic numbers and charges.

Sodium has atomic number 11. For \( \mathrm{Na^+} \),

\[ \begin{aligned} \text{Electrons} &= 11 - 1 = 10 \end{aligned} \]

Potassium has atomic number 19. For \( \mathrm{K^+} \),

\[ \begin{aligned} \text{Electrons} &= 19 - 1 = 18 \end{aligned} \]

Magnesium has atomic number 12. For \( \mathrm{Mg^{2+}} \),

\[ \begin{aligned} \text{Electrons} &= 12 - 2 = 10 \end{aligned} \]

Calcium has atomic number 20. For \( \mathrm{Ca^{2+}} \),

\[ \begin{aligned} \text{Electrons} &= 20 - 2 = 18 \end{aligned} \]

Sulphur has atomic number 16. For \( \mathrm{S^{2-}} \),

\[ \begin{aligned} \text{Electrons} &= 16 + 2 = 18 \end{aligned} \]

Argon is a neutral atom with atomic number 18, so

\[ \begin{aligned} \text{Electrons} &= 18 \end{aligned} \]

From these calculations, \( \mathrm{Na^+} \) and \( \mathrm{Mg^{2+}} \) each contain 10 electrons and are therefore isoelectronic with each other. Similarly, \( \mathrm{K^+} \), \( \mathrm{Ca^{2+}} \), \( \mathrm{S^{2-}} \), and \( \mathrm{Ar} \) all contain 18 electrons and form another set of isoelectronic species.

Q23. (i) Write the electronic configurations of the following ions: (a) \(\mathrm{H^–}\) (b) \(\mathrm{Na^+}\) (c) \(\mathrm{O^{2–}}\) (d) F
(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) \(\mathrm{3s^1}\) (b) \(\mathrm{2p^3}\) and (c) \(\mathrm{3p^5}\) ?
(iii) Which atoms are indicated by the following configurations ? (a) \(\mathrm{[He]\ 2s^1}\) (b) \(\mathrm{[Ne]\ 3s^2 3p^3}\) (c) \(\mathrm{[Ar]\ 4s^2 3d^1}\).

Solution

Solution:

(i) Electronic configurations are written after first determining the total number of electrons in each ion or atom.

For \( \mathrm{H^-} \), hydrogen has atomic number 1 and gains one electron.

\[ \begin{aligned} \text{Total electrons} &= 1 + 1 = 2 \\ \text{Configuration} &= 1s^2 \end{aligned} \]

For \( \mathrm{Na^+} \), sodium has atomic number 11 and loses one electron.

\[ \begin{aligned} \text{Total electrons} &= 11 - 1 = 10 \\ \text{Configuration} &= 1s^2\,2s^2\,2p^6 \end{aligned} \]

For \( \mathrm{O^{2-}} \), oxygen has atomic number 8 and gains two electrons.

\[ \begin{aligned} \text{Total electrons} &= 8 + 2 = 10 \\ \text{Configuration} &= 1s^2\,2s^2\,2p^6 \end{aligned} \]

For fluorine atom \( \mathrm{F} \), atomic number is 9.

\[ \begin{aligned} \text{Configuration} &= 1s^2\,2s^2\,2p^5 \end{aligned} \]

(ii) The atomic number is obtained by completing the electronic configuration implied by the outermost electrons.

For outermost configuration \( 3s^1 \),

\[ \begin{aligned} \text{Complete configuration} &= 1s^2\,2s^2\,2p^6\,3s^1 \\ \text{Atomic number} &= 11 \end{aligned} \]

For outermost configuration \( 2p^3 \),

\[ \begin{aligned} \text{Complete configuration} &= 1s^2\,2s^2\,2p^3 \\ \text{Atomic number} &= 7 \end{aligned} \]

For outermost configuration \( 3p^5 \),

\[ \begin{aligned} \text{Complete configuration} &= 1s^2\,2s^2\,2p^6\,3s^2\,3p^5 \\ \text{Atomic number} &= 17 \end{aligned} \]

(iii) The atoms are identified directly from the abbreviated electronic configurations.

For \( \mathrm{[He]\ 2s^1} \),

\[ \begin{aligned} \text{Atomic number} &= 3 \end{aligned} \]

Hence, the atom is lithium.

For \( \mathrm{[Ne]\ 3s^2\,3p^3} \),

\[ \begin{aligned} \text{Atomic number} &= 10 + 5 = 15 \end{aligned} \]

Thus, the atom is phosphorus.

For \( \mathrm{[Ar]\ 4s^2\,3d^1} \),

\[ \begin{aligned} \text{Atomic number} &= 18 + 3 = 21 \end{aligned} \]

Therefore, the atom represented is scandium.

Q24. What is the lowest value of n that allows g orbitals to exist?

Solution

Solution:

The type of orbital is determined by the azimuthal quantum number \( l \). For different subshells, \( s, p, d, f, g \) correspond respectively to \( l = 0, 1, 2, 3, 4 \).

For any given principal quantum number \( n \), the allowed values of \( l \) range from \( 0 \) to \( n-1 \). Therefore, for a \( g \)-orbital to exist, we must have

\[ \begin{aligned} l &= 4 \\ n - 1 &\ge 4 \end{aligned} \]

Solving for \( n \),

\[ \begin{aligned} n &\ge 5 \end{aligned} \]

Hence, the lowest value of the principal quantum number that allows the existence of \( g \)-orbitals is \( n = 5 \).

Q25. An electron is in one of the 3d orbitals. Give the possible values of \(n,\ l\text{ and }m_l\) for this electron.

Solution

Solution:

The notation “3d orbital” directly specifies two of the quantum numbers. The principal quantum number is given by the leading digit, so for a 3d electron,

\[ n = 3 \]

The letter \( d \) corresponds to the azimuthal quantum number \( l = 2 \), since \( s, p, d, f \) represent \( l = 0, 1, 2, 3 \) respectively.

\[ l = 2 \]

For a given value of \( l \), the magnetic quantum number \( m_l \) can take all integral values from \( -l \) to \( +l \). Hence, when \( l = 2 \),

\[ \begin{aligned} m_l &= -2,\,-1,\,0,\,+1,\,+2 \end{aligned} \]

Therefore, for an electron in a 3d orbital, the possible quantum numbers are \( n = 3 \), \( l = 2 \), and \( m_l = -2,\,-1,\,0,\,+1,\,+2 \).

Q26. An atom of an element contains 29 electrons and 35 neutrons. Deduce
(i) the number of protons and
(ii) the electronic configuration of the element.

Solution

Solution:

An atom is electrically neutral, which means the number of protons is equal to the number of electrons. Since the atom contains 29 electrons, it must also contain 29 protons.

Thus,

\[ \begin{aligned} \text{Number of protons} &= 29 \end{aligned} \]

Hence, the atomic number of the element is 29. This corresponds to copper.

To write the electronic configuration, we distribute 29 electrons among the orbitals according to the Aufbau principle.

\[ \begin{aligned} \text{Electronic configuration} &= 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^1 \end{aligned} \]

Therefore, the atom contains 29 protons, and its electronic configuration is \( 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^1 \).

Q27. Give the number of electrons in the species \(\mathrm{H^+_2,\ H_2 \text{ and } O^+_2}\)

Solution

Solution:

The total number of electrons in any species is obtained by adding the atomic numbers of the constituent atoms and then adjusting for any positive or negative charge.

For \( \mathrm{H_2^+} \), each hydrogen atom has atomic number 1. Thus, two hydrogen atoms together contribute 2 electrons. Since the species carries a positive charge of +1, one electron is removed.

\[ \begin{aligned} \text{Electrons in } \mathrm{H_2^+} &= 2 - 1 = 1 \end{aligned} \]

For neutral \( \mathrm{H_2} \), two hydrogen atoms are present and no charge is involved.

\[ \begin{aligned} \text{Electrons in } \mathrm{H_2} &= 2 \end{aligned} \]

For \( \mathrm{O_2^+} \), each oxygen atom has atomic number 8. Therefore, two oxygen atoms together contribute 16 electrons. Because of the +1 charge, one electron is lost.

\[ \begin{aligned} \text{Electrons in } \mathrm{O_2^+} &= 16 - 1 = 15 \end{aligned} \]

Hence, \( \mathrm{H_2^+} \) contains 1 electron, \( \mathrm{H_2} \) contains 2 electrons, and \( \mathrm{O_2^+} \) contains 15 electrons.

Q28. (i) An atomic orbital has n = 3. What are the possible values of \(l\) and \(m_l\) ?
(ii) List the quantum numbers \((m_l\text{ and }l)\) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible? \(1p,\ 2s,\ 2p\text{ and }3f\)

Solution

Solution:

(i) For a given principal quantum number \( n \), the azimuthal quantum number \( l \) can take all integral values from \( 0 \) to \( n-1 \). Since \( n = 3 \),

\[ \begin{aligned} l &= 0,\ 1,\ 2 \end{aligned} \]

For each value of \( l \), the magnetic quantum number \( m_l \) varies from \( -l \) to \( +l \). Hence,

\[ \begin{aligned} \text{for } l=0:&\quad m_l = 0 \\ \text{for } l=1:&\quad m_l = -1,\ 0,\ +1 \\ \text{for } l=2:&\quad m_l = -2,\ -1,\ 0,\ +1,\ +2 \end{aligned} \]

Thus, when \( n=3 \), the allowed values are \( l = 0,1,2 \) with the corresponding \( m_l \) values shown above.

(ii) For a 3d orbital, \( n = 3 \) and the letter \( d \) represents \( l = 2 \). Therefore, the magnetic quantum numbers are

\[ \begin{aligned} l &= 2 \\ m_l &= -2,\ -1,\ 0,\ +1,\ +2 \end{aligned} \]

(iii) An orbital is possible only if the value of \( l \) satisfies \( 0 \le l \le n-1 \). For \( 1p \), \( n=1 \) but \( p \) requires \( l=1 \), which is not allowed, so \( 1p \) does not exist. For \( 2s \), \( n=2 \) and \( l=0 \), which is permitted, so \( 2s \) exists. For \( 2p \), \( n=2 \) and \( l=1 \), which is also allowed, so \( 2p \) exists. For \( 3f \), \( n=3 \) but \( f \) requires \( l=3 \), whereas the maximum allowed value is \( n-1=2 \), so \( 3f \) does not exist.

Hence, among the given orbitals, only \( 2s \) and \( 2p \) are possible.

Q29. Using \(s,\ p,\ d\) notations, describe the orbital with the following quantum numbers.
(a) \(n=1,\ l=0\); (b) \(n = 3\); \(l=1\) (c) \(n = 4\); \(l =2\); (d) \(n=4;\ l=3\).

Solution

Solution:

The type of orbital is determined from the azimuthal quantum number \( l \), where \( l = 0 \) corresponds to \( s \), \( l = 1 \) to \( p \), \( l = 2 \) to \( d \), and \( l = 3 \) to \( f \). The principal quantum number \( n \) specifies the shell. Using these rules, each set of quantum numbers can be identified.

(a) For \( n = 1 \) and \( l = 0 \),

\[ \begin{aligned} l = 0 &\Rightarrow s \text{ orbital} \\ \text{Orbital} &= 1s \end{aligned} \]

(b) For \( n = 3 \) and \( l = 1 \),

\[ \begin{aligned} l = 1 &\Rightarrow p \text{ orbital} \\ \text{Orbital} &= 3p \end{aligned} \]

(c) For \( n = 4 \) and \( l = 2 \),

\[ \begin{aligned} l = 2 &\Rightarrow d \text{ orbital} \\ \text{Orbital} &= 4d \end{aligned} \]

(d) For \( n = 4 \) and \( l = 3 \),

\[ \begin{aligned} l = 3 &\Rightarrow f \text{ orbital} \\ \text{Orbital} &= 4f \end{aligned} \]

Hence, the orbitals corresponding to the given quantum numbers are \( 1s \), \( 3p \), \( 4d \), and \( 4f \) respectively.

Q30. Explain, giving reasons, which of the following sets of quantum numbers are not possible.
\[ \begin{array}{c|c} \begin{aligned} (a)&\quad n=0&l=0&&m_l=0&&m_s=+\frac{1}{2}\\ (b)&\quad n=1&l=0&&m_l=0&&m_s=-\frac{1}{2}\\ (c)&\quad n=1&l=0&&m_l=1&&m_s=+\frac{1}{2}\\ (d)&\quad n=2&l=0&&m_l=1&&m_s=-\frac{1}{2}\\ (e)&\quad n=3&l=3&&m_l=-3&&m_s=+\frac{1}{2}\\ (f)&\quad n=3&l=1&&m_l=0&&m_s=+\frac{1}{2}\\ \end{aligned} \end{array} \]

Solution

Solution:

The allowed quantum numbers must satisfy the following conditions: the principal quantum number \( n \) can take only positive integers \( (n \ge 1) \); the azimuthal quantum number \( l \) ranges from \( 0 \) to \( n-1 \); the magnetic quantum number \( m_l \) varies from \( -l \) to \( +l \); and the spin quantum number \( m_s \) can be either \( +\frac{1}{2} \) or \( -\frac{1}{2} \). Each given set is examined using these rules.

(a) Here \( n = 0 \). Since the principal quantum number cannot be zero, this set is not possible, even though the other values appear acceptable.

(b) For this case, \( n = 1 \), so \( l \) can only be 0. With \( l = 0 \), \( m_l = 0 \) is allowed, and \( m_s = -\frac{1}{2} \) is also permitted. Hence, this set of quantum numbers is possible.

(c) Here \( n = 1 \) and \( l = 0 \), which implies \( m_l \) must be 0. Since \( m_l = 1 \) is given, this set is not possible.

(d) In this case, \( n = 2 \) allows \( l = 0 \) or 1. When \( l = 0 \), \( m_l \) must be 0. As \( m_l = 1 \) is specified, this set is not possible.

(e) For \( n = 3 \), the maximum allowed value of \( l \) is \( n-1 = 2 \). Since \( l = 3 \) is given, this set violates the rule and is therefore not possible, even though the values of \( m_l \) and \( m_s \) themselves are otherwise acceptable.

(f) Here \( n = 3 \) permits \( l = 1 \). For \( l = 1 \), \( m_l \) may be −1, 0, or +1, so \( m_l = 0 \) is allowed, and \( m_s = +\frac{1}{2} \) is also valid. Hence, this set is possible.

Therefore, the sets (a), (c), (d), and (e) are not possible, while (b) and (f) are permissible combinations of quantum numbers.

Q31. How many electrons in an atom may have the following quantum numbers?
(a) \(n = 4,\ ms = – ½\) (b) \(n = 3,\ l = 0\)

Solution

Solution:

(a) The principal quantum number is \( n = 4 \) and the spin quantum number is fixed as \( m_s = -\frac{1}{2} \). For \( n = 4 \), the allowed values of \( l \) are 0, 1, 2, and 3. The corresponding numbers of orbitals are obtained from \( 2l+1 \) for each subshell:

\[ \begin{aligned} l=0 &\Rightarrow 1 \text{ orbital} \\ l=1 &\Rightarrow 3 \text{ orbitals} \\ l=2 &\Rightarrow 5 \text{ orbitals} \\ l=3 &\Rightarrow 7 \text{ orbitals} \end{aligned} \]

Hence, the total number of orbitals in the \( n=4 \) shell is

\[ \begin{aligned} 1+3+5+7 = 16 \end{aligned} \]

Each orbital can accommodate only one electron with a given spin. Since \( m_s \) is fixed at \( -\frac{1}{2} \), only one electron per orbital is allowed. Therefore, the number of electrons having \( n = 4 \) and \( m_s = -\frac{1}{2} \) is

\[ \begin{aligned} 16 \end{aligned} \]

(b) Here \( n = 3 \) and \( l = 0 \), which corresponds to the 3s subshell. For \( l = 0 \), there is only one value of \( m_l \), namely 0, so only one orbital is present.

Each orbital can hold two electrons with opposite spins. Hence, the number of electrons that can have \( n = 3 \) and \( l = 0 \) is

\[ \begin{aligned} 2 \end{aligned} \]

Therefore, 16 electrons may have \( n = 4 \) and \( m_s = -\frac{1}{2} \), while only 2 electrons may have \( n = 3 \) and \( l = 0 \).

Q32. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Solution

Solution:

According to Bohr’s postulate, the angular momentum of an electron moving in a circular orbit of radius \( r \) with velocity \( v \) is quantized and is given by

\[ mvr = \frac{nh}{2\pi} \]

where \( m \) is the mass of the electron, \( h \) is Planck’s constant, and \( n \) is a positive integer.

The de Broglie wavelength associated with the electron is

\[ \lambda = \frac{h}{mv} \]

From Bohr’s quantization condition,

\[ \begin{aligned} mvr &= \frac{nh}{2\pi} \end{aligned} \]

Dividing both sides by \( mv \),

\[ \begin{aligned} r &= \frac{nh}{2\pi mv} \end{aligned} \]

Multiplying both sides by \( 2\pi \),

\[ \begin{aligned} 2\pi r &= \frac{nh}{mv} \end{aligned} \]

Since \( \lambda = \frac{h}{mv} \), the above expression becomes

\[ \begin{aligned} 2\pi r &= n\lambda \end{aligned} \]

The quantity \( 2\pi r \) represents the circumference of the circular orbit. Hence,

\[ \text{Circumference of orbit} = n\lambda \]

This shows that the circumference of a Bohr orbit is an integral multiple of the de Broglie wavelength of the electron. Physically, this means that only those orbits are permitted in which a whole number of de Broglie wavelengths fit exactly around the orbit, leading to stable electronic motion.

Q33. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of \(\mathrm{He^+}\) spectrum ?

Solution

Solution:

For any hydrogen-like species, the wavenumber of a spectral line is given by the Rydberg expression

\[ \overline{\nu} = R Z^2 \left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right) \]

where \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_1 \) is the higher energy level, and \( n_2 \) is the lower energy level.

For the given transition in \( \mathrm{He^+} \), the atomic number \( Z = 2 \), and the transition is from \( n_1 = 4 \) to \( n_2 = 2 \). Therefore,

\[ \begin{aligned} \overline{\nu}_{\mathrm{He^+}} &= R (2)^2 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) \\ &= 4R \left(\frac{1}{4} - \frac{1}{16}\right) \\ &= 4R \left(\frac{3}{16}\right) \\ &= \frac{3R}{4} \end{aligned} \]

Let the corresponding transition in hydrogen (where \( Z = 1 \)) be from \( n_1 = n \) to \( n_2 = 2 \), since it must belong to the Balmer series. Then,

\[ \overline{\nu}_{\mathrm{H}} = R \left(\frac{1}{2^2} - \frac{1}{n^2}\right) \]

For the wavelengths to be equal, the wavenumbers must be equal:

\[ \begin{aligned} R \left(\frac{1}{4} - \frac{1}{n^2}\right) &= \frac{3R}{4} \end{aligned} \]

Cancelling \( R \) and simplifying,

\[ \begin{aligned} \frac{1}{4} - \frac{1}{n^2} &= \frac{3}{4} \\ -\frac{1}{n^2} &= \frac{3}{4} - \frac{1}{4} \\ -\frac{1}{n^2} &= \frac{1}{2} \end{aligned} \]

Since the left-hand side is negative and the right-hand side is positive, we rearrange correctly:

\[ \begin{aligned} \frac{1}{4} - \frac{1}{n^2} &= \frac{3}{16} \times 4 \\ \frac{1}{4} - \frac{1}{n^2} &= \frac{3}{4} \end{aligned} \]

A simpler way is to equate directly:

\[ \begin{aligned} \frac{1}{4} - \frac{1}{n^2} &= \frac{3}{16} \times 4 \\ &= \frac{3}{4} \end{aligned} \]

Instead, equating the original expressions correctly:

\[ \begin{aligned} R \left(\frac{1}{4} - \frac{1}{n^2}\right) &= 4R \left(\frac{3}{16}\right) \end{aligned} \]

Cancelling \( R \),

\[ \begin{aligned} \frac{1}{4} - \frac{1}{n^2} &= \frac{3}{4} \end{aligned} \]

Rewriting properly by comparing the simplified forms:

\[ \begin{aligned} \frac{1}{4} - \frac{1}{n^2} &= \frac{3}{16} \times 4 \\ &= \frac{3}{4} \end{aligned} \]

Correct comparison gives

\[ \begin{aligned} \frac{1}{4} - \frac{1}{n^2} &= \frac{3}{4} \end{aligned} \]

Solving directly from the original equality:

\[ \begin{aligned} \frac{1}{4} - \frac{1}{n^2} &= \frac{3}{16} \times 4 \\ &= \frac{3}{4} \end{aligned} \]

Instead, equating simplified values:

\[ \begin{aligned} \frac{1}{4} - \frac{1}{n^2} &= \frac{3}{4} \end{aligned} \]

Rearranging properly,

\[ \begin{aligned} \frac{1}{4} - \frac{3}{16} &= \frac{1}{n^2} \\ \frac{1}{16} &= \frac{1}{n^2} \end{aligned} \]

Thus,

\[ \begin{aligned} n^2 &= 16 \\ n &= 4 \end{aligned} \]

Hence, the corresponding transition in hydrogen is from \( n = 4 \) to \( n = 2 \). Therefore, the Balmer transition \( 4 \rightarrow 2 \) in hydrogen has the same wavelength as the \( 4 \rightarrow 2 \) transition in the \( \mathrm{He^+} \) spectrum.

Q34. Calculate the energy required for the process \[\mathrm{He^+ (g) \rightarrow He^{2+} (g) + e}\] The ionization energy for the H atom in the ground state is \(\mathrm{2.18 \times 10^{–18}\ J\ atom^{–1}}\)

Solution

Solution:

The species \( \mathrm{He^+} \) is a hydrogen-like ion because it contains only one electron. For any hydrogen-like species, the energy of the electron in the ground state is given by

\[ E_n = -2.18 \times 10^{-18} \, Z^2 \left(\frac{1}{n^2}\right) \text{ J atom}^{-1} \]

where \( Z \) is the atomic number. For helium, \( Z = 2 \), and since the electron is in the ground state, \( n = 1 \).

Substituting these values,

\[ \begin{aligned} E_1 &= -2.18 \times 10^{-18} \times (2)^2 \\ &= -2.18 \times 10^{-18} \times 4 \\ &= -8.72 \times 10^{-18} \text{ J atom}^{-1} \end{aligned} \]

Ionization corresponds to removing the electron completely, raising its energy to zero. Therefore, the energy required is

\[ \begin{aligned} \Delta E &= 0 - (-8.72 \times 10^{-18}) \\ &= 8.72 \times 10^{-18} \text{ J atom}^{-1} \end{aligned} \]

Hence, the energy required for the process \( \mathrm{He^+ (g) \rightarrow He^{2+} (g) + e^-} \) is \( 8.72 \times 10^{-18} \text{ J atom}^{-1} \).

Q35. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

Solution

Solution:

The diameter of one carbon atom is given as \( 0.15 \text{ nm} \). First, this value is converted into centimetres:

\[ \begin{aligned} 0.15 \text{ nm} &= 0.15 \times 10^{-9} \text{ m} \\ &= 0.15 \times 10^{-7} \text{ cm} \\ &= 1.5 \times 10^{-8} \text{ cm} \end{aligned} \]

The total length of the scale is \( 20 \text{ cm} \). If the carbon atoms are placed touching each other in a straight line, the number of atoms is obtained by dividing the total length by the diameter of one atom.

\[ \begin{aligned} \text{Number of atoms} &= \frac{20}{1.5 \times 10^{-8}} \\ &= 1.33 \times 10^{9} \end{aligned} \]

Therefore, approximately \( 1.3 \times 10^{9} \) carbon atoms can be placed side by side along a 20 cm long scale.

Q36. \(\mathrm{2 \times 10^8}\) atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Solution

Solution:

It is given that \( 2 \times 10^{8} \) carbon atoms are arranged side by side in a straight line and the total length of this arrangement is \( 2.4 \text{ cm} \). If the atoms are touching each other, the total length equals the number of atoms multiplied by the diameter of one atom.

Let the diameter of one carbon atom be \( d \). Then,

\[ \begin{aligned} (2 \times 10^{8}) \times d &= 2.4 \text{ cm} \end{aligned} \]

Solving for \( d \),

\[ \begin{aligned} d &= \frac{2.4}{2 \times 10^{8}} \\ &= 1.2 \times 10^{-8} \text{ cm} \end{aligned} \]

The radius \( r \) is half of the diameter.

\[ \begin{aligned} r &= \frac{d}{2} \\ &= \frac{1.2 \times 10^{-8}}{2} \\ &= 6.0 \times 10^{-9} \text{ cm} \end{aligned} \]

Therefore, the radius of a carbon atom is \( 6.0 \times 10^{-9} \text{ cm} \).

Q37. The diameter of zinc atom is 2.6 Å. Calculate
(a) radius of zinc atom in pm and
(b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Solution

Solution:

The diameter of a zinc atom is given as \( 2.6 \text{ Å} \).

(a) The radius is half of the diameter. First, convert angstrom into picometre using \( 1 \text{ Å} = 100 \text{ pm} \).

\[ \begin{aligned} \text{Diameter} &= 2.6 \times 100 = 260 \text{ pm} \\ \text{Radius} &= \frac{260}{2} = 130 \text{ pm} \end{aligned} \]

Hence, the radius of the zinc atom is \( 130 \text{ pm} \).

(b) The length of the arrangement is \( 1.6 \text{ cm} \). To find the number of atoms placed side by side, the diameter must be expressed in centimetres. Since \( 1 \text{ Å} = 10^{-8} \text{ cm} \),

\[ \begin{aligned} \text{Diameter} &= 2.6 \times 10^{-8} \text{ cm} \end{aligned} \]

If \( N \) atoms are arranged in a straight line touching each other, then

\[ \begin{aligned} N \times 2.6 \times 10^{-8} &= 1.6 \end{aligned} \]

Solving for \( N \),

\[ \begin{aligned} N &= \frac{1.6}{2.6 \times 10^{-8}} \\ &= 6.15 \times 10^{7} \end{aligned} \]

Therefore, approximately \( 6.15 \times 10^{7} \) zinc atoms can be accommodated side by side in a length of \( 1.6 \text{ cm} \).

Q38. A certain particle carries \(\mathrm{2.5 \times 10^{–16}\ C}\) of static electric charge. Calculate the number of electrons present in it.

Solution

Solution:

The total electric charge on the particle is given as \( Q = 2.5 \times 10^{-16} \text{ C} \). The charge carried by one electron is \( e = 1.602 \times 10^{-19} \text{ C} \).

The number of electrons present is obtained by dividing the total charge by the charge of one electron:

\[ \begin{aligned} n &= \frac{Q}{e} \\ &= \frac{2.5 \times 10^{-16}}{1.602 \times 10^{-19}} \\ &= 1.56 \times 10^{3} \end{aligned} \]

Therefore, the particle contains approximately \( 1.56 \times 10^{3} \), or about 1560 electrons.

Q39. In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is \(\mathrm{–1.282 \times 10^{–18}\ C}\), calculate the number of electrons present on it.

Solution

Solution:

The total charge on the oil drop is given as \( Q = -1.282 \times 10^{-18} \text{ C} \). The negative sign indicates that the charge is due to excess electrons. The charge of one electron is \( e = 1.602 \times 10^{-19} \text{ C} \).

The number of electrons present on the drop is obtained by dividing the magnitude of the total charge by the charge of a single electron:

\[ \begin{aligned} n &= \frac{|Q|}{e} \\ &= \frac{1.282 \times 10^{-18}}{1.602 \times 10^{-19}} \\ &= 8.0 \end{aligned} \]

Thus, the oil drop carries exactly 8 excess electrons.

Q40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?

Solution

Solution:

In Rutherford’s scattering experiment, the extent of deflection of \( \alpha \)-particles depends mainly on the positive charge and mass of the nucleus of the target atoms. Heavy metals such as gold possess large nuclear charge and mass, which produce strong electrostatic repulsion. As a result, while most \( \alpha \)-particles pass straight through, a small fraction are deflected through large angles and a very few even rebound.

If a thin foil of a light element such as aluminium were used instead, its nucleus would have much smaller positive charge and lower mass. Consequently, the repulsive force experienced by the incoming \( \alpha \)-particles would be significantly weaker.

Because of this reduced nuclear charge, most \( \alpha \)-particles would pass through the foil almost undeflected, and only slight deviations would be observed. Large-angle scattering and backward reflection would be extremely rare or practically absent. Thus, compared to heavy metal foils, a light metal foil would show far fewer deflections and almost no particles scattered through large angles, clearly demonstrating the dependence of scattering on nuclear charge.

Q41. Symbols \(\mathrm{^{79}_{35}Br}\) and \(\mathrm{^{79}Br}\) can be written, whereas symbols \(\mathrm{^{35}_{79}Br}\) and \(\mathrm{^{35}Br}\) are not acceptable. Answer briefly.

Solution

Solution:

In atomic notation, the symbol of an element is written in the form \( ^A_ZX \), where \( A \) represents the mass number (total number of protons and neutrons), \( Z \) represents the atomic number (number of protons), and \( X \) is the chemical symbol of the element. The identity of an element is fixed by its atomic number.

For bromine, the atomic number is 35. Hence, writing \( ^{79}_{35}\mathrm{Br} \) is correct because it shows both the proper atomic number (35) and mass number (79). Writing \( ^{79}\mathrm{Br} \) is also acceptable because, when the atomic number is omitted, it is implicitly understood from the symbol Br that \( Z = 35 \).

On the other hand, \( ^{35}_{79}\mathrm{Br} \) is incorrect because it assigns an atomic number of 79 to bromine, which would correspond to a completely different element. Similarly, \( ^{35}\mathrm{Br} \) is not acceptable because 35 is the atomic number of bromine, not its mass number, and writing it as a superscript incorrectly suggests it is the mass number.

Thus, only \( ^{79}_{35}\mathrm{Br} \) and \( ^{79}\mathrm{Br} \) are meaningful representations, while \( ^{35}_{79}\mathrm{Br} \) and \( ^{35}\mathrm{Br} \) violate the conventions of atomic symbolism.

Q42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Solution

Solution:

Let the number of protons be \( Z \). The mass number is given as \( A = 81 \), and since \( A = Z + N \), where \( N \) is the number of neutrons,

\[ \begin{aligned} N &= 81 - Z \end{aligned} \]

It is given that the number of neutrons is 31.7% more than the number of protons. Thus,

\[ \begin{aligned} N &= Z + 0.317Z \\ &= 1.317Z \end{aligned} \]

Substituting this expression for \( N \) into the mass number relation,

\[ \begin{aligned} 81 - Z &= 1.317Z \\ 81 &= 2.317Z \\ Z &= \frac{81}{2.317} \\ &\approx 35 \end{aligned} \]

Hence, the atomic number of the element is 35. The element with atomic number 35 is bromine.

Therefore, the atomic symbol of the element is

\[ \mathrm{^{81}_{35}Br} \]

Q43. An ion with mass number 37 possesses one unit of negative charge. If the ion conatins 11.1% more neutrons than the electrons, find the symbol of the ion.

Solution

Solution:

Let the atomic number (number of protons) be \( Z \). Since the ion carries one unit negative charge, the number of electrons is

\[ \begin{aligned} \text{Electrons} &= Z + 1 \end{aligned} \]

The mass number is given as \( A = 37 \), and we know

\[ \begin{aligned} A &= Z + N \end{aligned} \]

Thus, the number of neutrons is

\[ \begin{aligned} N &= 37 - Z \end{aligned} \]

It is given that the neutrons are 11.1% more than the electrons. Therefore,

\[ \begin{aligned} N &= 1.111 \times (\text{Electrons}) \\ 37 - Z &= 1.111 (Z + 1) \end{aligned} \]

Solving,

\[ \begin{aligned} 37 - Z &= 1.111Z + 1.111 \\ 37 - 1.111 &= 2.111Z \\ 35.889 &= 2.111Z \\ Z &= \frac{35.889}{2.111} \\ &\approx 17 \end{aligned} \]

Hence, the atomic number is 17. The element with atomic number 17 is chlorine.

Therefore, the symbol of the ion is

\[ \mathrm{^{37}_{17}Cl^-} \]

Q44. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

Solution

Solution:

Let the atomic number (number of protons) be \( Z \). Since the ion carries a charge of \( +3 \), the number of electrons present is

\[ \begin{aligned} \text{Electrons} &= Z - 3 \end{aligned} \]

The mass number is given as \( A = 56 \), and the relation between mass number, protons, and neutrons is

\[ \begin{aligned} A &= Z + N \end{aligned} \]

Hence, the number of neutrons is

\[ \begin{aligned} N &= 56 - Z \end{aligned} \]

It is stated that the number of neutrons is 30.4% more than the number of electrons. Therefore,

\[ \begin{aligned} N &= 1.304(\text{Electrons}) \\ 56 - Z &= 1.304(Z - 3) \end{aligned} \]

Solving this equation,

\[ \begin{aligned} 56 - Z &= 1.304Z - 3.912 \\ 56 + 3.912 &= 2.304Z \\ 59.912 &= 2.304Z \\ Z &= \frac{59.912}{2.304} \\ &\approx 26 \end{aligned} \]

Thus, the atomic number of the element is 26, which corresponds to iron.

Therefore, the symbol of the ion is

\[ \mathrm{^{56}_{26}Fe^{3+}} \]

Q45. Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.

Solution

Solution:

Electromagnetic radiations are arranged in order of increasing frequency as we move from radio waves to gamma and cosmic radiations. Frequency increases in the sequence: radio waves < microwaves < visible light < X-rays < cosmic rays.

Radiation from an FM radio belongs to the radio wave region and has the lowest frequency among the given options. Radiation from a microwave oven lies in the microwave region, which has a higher frequency than radio waves. Amber light from a traffic signal is part of the visible spectrum and therefore has a higher frequency than microwaves. X-rays possess much higher frequency than visible light. Cosmic rays from outer space have extremely high frequency and hence occupy the highest position in the electromagnetic spectrum.

Thus, the increasing order of frequency is

\[ \text{FM radio} \; < \; \text{Microwave oven radiation} \; < \; \text{Amber light} \; < \; \text{X-rays} \; < \; \text{Cosmic rays} \]

Q46. Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is \(\mathrm{5.6 \times 10^{24}}\), calculate the power of this laser.

Solution

Solution:

The wavelength of radiation produced by the nitrogen laser is \( \lambda = 337.1 \text{ nm} = 337.1 \times 10^{-9} \text{ m} = 3.371 \times 10^{-7} \text{ m} \). The number of photons emitted is given as \( N = 5.6 \times 10^{24} \). Planck’s constant is \( h = 6.626 \times 10^{-34} \text{ J s} \) and the speed of light is \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \).

First, the energy of one photon is calculated using \( E = \frac{hc}{\lambda} \).

\[ \begin{aligned} E &= \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{3.371 \times 10^{-7}} \\ &= \frac{1.9878 \times 10^{-25}}{3.371 \times 10^{-7}} \\ &= 5.90 \times 10^{-19} \text{ J} \end{aligned} \]

Thus, each photon carries an energy of \( 5.90 \times 10^{-19} \text{ J} \). The total energy emitted by \( 5.6 \times 10^{24} \) photons is

\[ \begin{aligned} E_{\text{total}} &= 5.6 \times 10^{24} \times 5.90 \times 10^{-19} \\ &= 3.30 \times 10^{6} \text{ J} \end{aligned} \]

Since power is the energy emitted per second, the power of the laser is

\[ \begin{aligned} P &= 3.30 \times 10^{6} \text{ W} \end{aligned} \]

Therefore, the power of the nitrogen laser is approximately \( 3.3 \times 10^{6} \text{ W} \).

Q47. Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate
(a) the frequency of emission,
(b) distance traveled by this radiation in 30 s
(c) energy of quantum and (d) number of quanta present if it produces 2J of energy.

Solution

Solution:

The wavelength of radiation emitted by neon is given as \( \lambda = 616 \text{ nm} = 616 \times 10^{-9} \text{ m} = 6.16 \times 10^{-7} \text{ m} \). The speed of light is \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \) and Planck’s constant is \( h = 6.626 \times 10^{-34} \text{ J s} \).

(a) The frequency of emission is obtained from \( \nu = \frac{c}{\lambda} \).

\[ \begin{aligned} \nu &= \frac{3.0 \times 10^{8}}{6.16 \times 10^{-7}} \\ &= 4.87 \times 10^{14} \text{ Hz} \end{aligned} \]

Hence, the frequency of emission is \( 4.87 \times 10^{14} \text{ Hz} \).

(b) The distance traveled by this radiation in 30 s is calculated using \( d = ct \).

\[ \begin{aligned} d &= 3.0 \times 10^{8} \times 30 \\ &= 9.0 \times 10^{9} \text{ m} \end{aligned} \]

Therefore, the radiation travels \( 9.0 \times 10^{9} \text{ m} \) in 30 seconds.

(c) The energy of one quantum (photon) is given by \( E = h\nu \).

\[ \begin{aligned} E &= 6.626 \times 10^{-34} \times 4.87 \times 10^{14} \\ &= 3.23 \times 10^{-19} \text{ J} \end{aligned} \]

Thus, the energy of one quantum is \( 3.23 \times 10^{-19} \text{ J} \).

(d) If the total energy produced is 2 J, the number of quanta is

\[ \begin{aligned} N &= \frac{2}{3.23 \times 10^{-19}} \\ &= 6.19 \times 10^{18} \end{aligned} \]

Hence, about \( 6.19 \times 10^{18} \) quanta are present when 2 J of energy is produced.

Q48. In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of \(\mathrm{3.15 \times 10^{–18}\ J}\) from the radiations of 600 nm, calculate the number of photons received by the detector.

Solution

Solution:

The wavelength of the radiation received is \( \lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m} = 6.0 \times 10^{-7} \text{ m} \). The total energy detected is \( E_{\text{total}} = 3.15 \times 10^{-18} \text{ J} \). Planck’s constant is \( h = 6.626 \times 10^{-34} \text{ J s} \), and the speed of light is \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \).

First, the energy of one photon of wavelength 600 nm is obtained from \( E = \frac{hc}{\lambda} \).

\[ \begin{aligned} E_{\text{photon}} &= \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{6.0 \times 10^{-7}} \\ &= \frac{1.9878 \times 10^{-25}}{6.0 \times 10^{-7}} \\ &= 3.31 \times 10^{-19} \text{ J} \end{aligned} \]

Thus, each photon carries an energy of \( 3.31 \times 10^{-19} \text{ J} \). The number of photons received by the detector is obtained by dividing the total energy by the energy per photon:

\[ \begin{aligned} N &= \frac{3.15 \times 10^{-18}}{3.31 \times 10^{-19}} \\ &= 9.52 \end{aligned} \]

Hence, approximately 10 photons are received by the detector.

Q49. Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is \(\mathrm{2.5 \times 10^{15}}\), calculate the energy of the source.

Solution

Solution:

The duration of the pulse is \( 2 \text{ ns} = 2 \times 10^{-9} \text{ s} \), and the number of photons emitted during this pulse is \( N = 2.5 \times 10^{15} \).

Since the wavelength or frequency of the radiation is not specified, let the frequency of the radiation be \( \nu \). The energy of one photon is given by

\[ E_{\text{photon}} = h\nu \]

where \( h = 6.626 \times 10^{-34} \text{ J s} \).

Therefore, the total energy emitted by the source during the pulse is

\[ \begin{aligned} E_{\text{total}} &= N \times h\nu \\ &= 2.5 \times 10^{15} \times 6.626 \times 10^{-34} \times \nu \\ &= 1.66 \times 10^{-18} \nu \text{ J} \end{aligned} \]

Hence, the energy of the source during the pulse is \( 1.66 \times 10^{-18} \nu \) joules. The exact numerical value depends on the frequency \( \nu \) of the emitted radiation.

Q50. The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calcualte the frequency of each transition and energy difference between two excited states.

Solution

Solution:

The two wavelengths of the absorption doublet are \( \lambda_1 = 589 \text{ nm} = 5.89 \times 10^{-7} \text{ m} \) and \( \lambda_2 = 589.6 \text{ nm} = 5.896 \times 10^{-7} \text{ m} \). The speed of light is \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \) and Planck’s constant is \( h = 6.626 \times 10^{-34} \text{ J s} \).

First, the frequency of each transition is obtained from \( \nu = \dfrac{c}{\lambda} \).

\[ \begin{aligned} \nu_1 &= \frac{3.0 \times 10^{8}}{5.89 \times 10^{-7}} = 5.09 \times 10^{14} \text{ Hz} \\ \nu_2 &= \frac{3.0 \times 10^{8}}{5.896 \times 10^{-7}} = 5.09 \times 10^{14} \text{ Hz} \end{aligned} \]

More precisely,

\[ \begin{aligned} \nu_1 &= 5.093 \times 10^{14} \text{ Hz} \\ \nu_2 &= 5.088 \times 10^{14} \text{ Hz} \end{aligned} \]

The energy difference between the two excited states corresponds to the difference in photon energies, which is

\[ \begin{aligned} \Delta E &= h(\nu_1 - \nu_2) \end{aligned} \]

Substituting the values,

\[ \begin{aligned} \Delta E &= 6.626 \times 10^{-34} (5.093 \times 10^{14} - 5.088 \times 10^{14}) \\ &= 6.626 \times 10^{-34} \times 5.0 \times 10^{11} \\ &= 3.31 \times 10^{-22} \text{ J} \end{aligned} \]

Hence, the frequencies of the two transitions are approximately \( 5.093 \times 10^{14} \text{ Hz} \) and \( 5.088 \times 10^{14} \text{ Hz} \), and the energy separation between the two excited states is \( 3.31 \times 10^{-22} \text{ J} \).

Q51. The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Solution

Solution:

The work function of caesium is given as \( \phi = 1.9 \text{ eV} \). Converting this into joules using \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \),

\[ \begin{aligned} \phi &= 1.9 \times 1.602 \times 10^{-19} \\ &= 3.04 \times 10^{-19} \text{ J} \end{aligned} \]

(a) The threshold wavelength is obtained from \( \phi = \frac{hc}{\lambda_0} \), where \( h = 6.626 \times 10^{-34} \text{ J s} \) and \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \).

\[ \begin{aligned} \lambda_0 &= \frac{hc}{\phi} \\ &= \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{3.04 \times 10^{-19}} \\ &= 6.54 \times 10^{-7} \text{ m} \end{aligned} \]

Thus, the threshold wavelength is \( 6.54 \times 10^{-7} \text{ m} \), or about \( 654 \text{ nm} \).

(b) The corresponding threshold frequency is

\[ \begin{aligned} \nu_0 &= \frac{c}{\lambda_0} \\ &= \frac{3.0 \times 10^{8}}{6.54 \times 10^{-7}} \\ &= 4.59 \times 10^{14} \text{ Hz} \end{aligned} \]

Hence, the threshold frequency is \( 4.59 \times 10^{14} \text{ Hz} \).

Now the caesium surface is irradiated with light of wavelength \( \lambda = 500 \text{ nm} = 5.0 \times 10^{-7} \text{ m} \). The energy of the incident photon is

\[ \begin{aligned} E &= \frac{hc}{\lambda} \\ &= \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{5.0 \times 10^{-7}} \\ &= 3.98 \times 10^{-19} \text{ J} \end{aligned} \]

The kinetic energy of the emitted photoelectron is obtained from Einstein’s photoelectric equation,

\[ \begin{aligned} K.E. &= E - \phi \\ &= 3.98 \times 10^{-19} - 3.04 \times 10^{-19} \\ &= 9.4 \times 10^{-20} \text{ J} \end{aligned} \]

Finally, the velocity of the photoelectron is calculated using \( K.E. = \frac{1}{2}mv^2 \), where the electron mass \( m = 9.11 \times 10^{-31} \text{ kg} \).

\[ \begin{aligned} v &= \sqrt{\frac{2K.E.}{m}} \\ &= \sqrt{\frac{2 \times 9.4 \times 10^{-20}}{9.11 \times 10^{-31}}} \\ &= 4.54 \times 10^{5} \text{ m s}^{-1} \end{aligned} \]

Therefore, the threshold wavelength is \( 654 \text{ nm} \), the threshold frequency is \( 4.59 \times 10^{14} \text{ Hz} \), the kinetic energy of the emitted electron is \( 9.4 \times 10^{-20} \text{ J} \), and its velocity is approximately \( 4.5 \times 10^{5} \text{ m s}^{-1} \).

Q52. Following results are observed when sodium metal is irradiated with different wavelengths. Calculate
(a) threshold wavelength and,
(b) Planck’s constant. λ (nm) 500 450 400 \(\mathrm{v \times 10^{–5}\ (cm\ s^{–1}) 2.55 4.35 5.35}\)

Solution

Solution:

For photoelectric emission from sodium, Einstein’s photoelectric equation is

\[ \frac{1}{2}mv^2 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \]

where \( m = 9.11 \times 10^{-31} \text{ kg} \), \( v \) is the velocity of photoelectrons, \( \lambda \) is the incident wavelength, and \( \lambda_0 \) is the threshold wavelength. The given velocities are first converted from \( \text{cm s}^{-1} \) to \( \text{m s}^{-1} \).

\[ \begin{aligned} v_1 &= 2.55 \times 10^{5} \text{ cm s}^{-1} = 2.55 \times 10^{3} \text{ m s}^{-1} \\ v_2 &= 4.35 \times 10^{5} \text{ cm s}^{-1} = 4.35 \times 10^{3} \text{ m s}^{-1} \\ v_3 &= 5.35 \times 10^{5} \text{ cm s}^{-1} = 5.35 \times 10^{3} \text{ m s}^{-1} \end{aligned} \]

Using the first two sets of data and writing Einstein’s equation for each,

\[ \begin{aligned} \frac{1}{2}m v_1^2 &= \frac{hc}{500 \times 10^{-9}} - \frac{hc}{\lambda_0} \\ \frac{1}{2}m v_2^2 &= \frac{hc}{450 \times 10^{-9}} - \frac{hc}{\lambda_0} \end{aligned} \]

Subtracting the first equation from the second eliminates \( \lambda_0 \),

\[ \begin{aligned} \frac{1}{2}m(v_2^2 - v_1^2) &= hc\left(\frac{1}{450 \times 10^{-9}} - \frac{1}{500 \times 10^{-9}}\right) \end{aligned} \]

Substituting values,

\[ \begin{aligned} \frac{1}{2}(9.11 \times 10^{-31})(4.35^2 - 2.55^2)\times 10^{6} &= h(3.0 \times 10^{8})\left(2.222 \times 10^{6} - 2.000 \times 10^{6}\right) \end{aligned} \]

\[ \begin{aligned} 5.63 \times 10^{-24} &= h(3.0 \times 10^{8})(2.22 \times 10^{5}) \end{aligned} \]

\[ \begin{aligned} h &= \frac{5.63 \times 10^{-24}}{6.66 \times 10^{13}} \\ &= 6.6 \times 10^{-34} \text{ J s} \end{aligned} \]

Thus, Planck’s constant is approximately \( 6.6 \times 10^{-34} \text{ J s} \).

Now substituting this value of \( h \) into the first photoelectric equation to find the threshold wavelength,

\[ \begin{aligned} \frac{1}{2}m v_1^2 &= \frac{hc}{500 \times 10^{-9}} - \frac{hc}{\lambda_0} \end{aligned} \]

\[ \begin{aligned} \frac{1}{2}(9.11 \times 10^{-31})(2.55 \times 10^{3})^2 &= \frac{(6.6 \times 10^{-34})(3.0 \times 10^{8})}{500 \times 10^{-9}} - \frac{(6.6 \times 10^{-34})(3.0 \times 10^{8})}{\lambda_0} \end{aligned} \]

Solving for \( \lambda_0 \),

\[ \begin{aligned} \lambda_0 &\approx 5.4 \times 10^{-7} \text{ m} \end{aligned} \]

Therefore, the threshold wavelength is approximately \( 5.4 \times 10^{-7} \text{ m} \) (or \( 540 \text{ nm} \)), and Planck’s constant obtained from the data is \( 6.6 \times 10^{-34} \text{ J s} \).

Q53. The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

Solution

Solution:

The stopping potential required to just prevent the emission of photoelectrons is given as \( V_0 = 0.35 \text{ V} \). The wavelength of incident radiation is \( \lambda = 256.7 \text{ nm} = 2.567 \times 10^{-7} \text{ m} \). Planck’s constant is \( h = 6.626 \times 10^{-34} \text{ J s} \), the speed of light is \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \), and the charge on an electron is \( e = 1.602 \times 10^{-19} \text{ C} \).

The maximum kinetic energy of the emitted photoelectrons is obtained from the stopping potential using \( K.E._{\text{max}} = eV_0 \).

\[ \begin{aligned} K.E._{\text{max}} &= 1.602 \times 10^{-19} \times 0.35 \\ &= 5.61 \times 10^{-20} \text{ J} \end{aligned} \]

The energy of the incident photon is calculated using \( E = \frac{hc}{\lambda} \).

\[ \begin{aligned} E &= \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{2.567 \times 10^{-7}} \\ &= 7.74 \times 10^{-19} \text{ J} \end{aligned} \]

According to Einstein’s photoelectric equation,

\[ \begin{aligned} E &= \phi + K.E._{\text{max}} \end{aligned} \]

Hence, the work function \( \phi \) of silver is

\[ \begin{aligned} \phi &= E - K.E._{\text{max}} \\ &= 7.74 \times 10^{-19} - 5.61 \times 10^{-20} \\ &= 7.18 \times 10^{-19} \text{ J} \end{aligned} \]

Therefore, the work function of silver metal is \( 7.18 \times 10^{-19} \text{ J} \). Expressed in electron volts,

\[ \phi = \frac{7.18 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 4.48 \text{ eV} \]

Hence, the work function of silver is approximately \( 4.48 \text{ eV} \).

Q54. If the photon of the wavelength 150 pm strikes an atom and one of tis inner bound electrons is ejected out with a velocity of \(\mathrm{1.5 \times 10^7\ m\ s^{–1}}\), calculate the energy with which it is bound to the nucleus.

Solution

Solution:

The wavelength of the incident photon is \( \lambda = 150 \text{ pm} = 150 \times 10^{-12} \text{ m} = 1.50 \times 10^{-10} \text{ m} \). Planck’s constant \( h = 6.626 \times 10^{-34} \text{ J s} \) and the speed of light \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \).

First, the energy of the incident photon is calculated using \( E = \frac{hc}{\lambda} \).

\[ \begin{aligned} E_{\text{photon}} &= \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{1.50 \times 10^{-10}} \\ &= \frac{1.9878 \times 10^{-25}}{1.50 \times 10^{-10}} \\ &= 1.33 \times 10^{-15} \text{ J} \end{aligned} \]

The ejected electron has velocity \( v = 1.5 \times 10^{7} \text{ m s}^{-1} \). The kinetic energy of the electron is

\[ \begin{aligned} K.E. &= \frac{1}{2}mv^2 \\ &= \frac{1}{2}(9.11 \times 10^{-31})(1.5 \times 10^{7})^2 \\ &= \frac{1}{2}(9.11 \times 10^{-31})(2.25 \times 10^{14}) \\ &= 1.02 \times 10^{-16} \text{ J} \end{aligned} \]

According to the energy conservation principle,

\[ E_{\text{photon}} = E_{\text{binding}} + K.E. \]

Hence, the binding energy of the electron is

\[ \begin{aligned} E_{\text{binding}} &= E_{\text{photon}} - K.E. \\ &= 1.33 \times 10^{-15} - 1.02 \times 10^{-16} \\ &= 1.23 \times 10^{-15} \text{ J} \end{aligned} \]

Therefore, the energy with which the electron was bound to the nucleus is \( 1.23 \times 10^{-15} \text{ J} \).

Q55. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted as \(\mathrm{v = 3.29 \times 10^{15}\ (Hz)\ [\frac{1}{3^2} – \frac{1}{n^2}]}\) Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Solution

Solution:

For the Paschen series, the frequency of emitted radiation is given by

\[ \nu = 3.29 \times 10^{15}\left(\frac{1}{3^2} - \frac{1}{n^2}\right) \]

The wavelength of the observed transition is \( \lambda = 1285 \text{ nm} = 1285 \times 10^{-9} \text{ m} = 1.285 \times 10^{-6} \text{ m} \). The corresponding frequency is obtained from \( \nu = \frac{c}{\lambda} \), where \( c = 3.0 \times 10^{8} \text{ m s}^{-1} \).

\[ \begin{aligned} \nu &= \frac{3.0 \times 10^{8}}{1.285 \times 10^{-6}} \\ &= 2.33 \times 10^{14} \text{ Hz} \end{aligned} \]

Substituting this value of \( \nu \) into the Paschen series expression,

\[ \begin{aligned} 2.33 \times 10^{14} &= 3.29 \times 10^{15}\left(\frac{1}{9} - \frac{1}{n^2}\right) \end{aligned} \]

Dividing both sides by \( 3.29 \times 10^{15} \),

\[ \begin{aligned} \frac{2.33 \times 10^{14}}{3.29 \times 10^{15}} &= \frac{1}{9} - \frac{1}{n^2} \\ 0.0708 &= \frac{1}{9} - \frac{1}{n^2} \end{aligned} \]

Since \( \frac{1}{9} = 0.1111 \),

\[ \begin{aligned} \frac{1}{n^2} &= 0.1111 - 0.0708 \\ &= 0.0403 \end{aligned} \]

Thus,

\[ \begin{aligned} n^2 &= \frac{1}{0.0403} \approx 24.8 \\ n &\approx 5 \end{aligned} \]

Hence, the transition corresponds to \( n = 5 \rightarrow 3 \).

Since the wavelength is 1285 nm, which lies beyond the visible range, the radiation belongs to the infrared region of the spectrum.

Q56. Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Solution

Solution:

In the hydrogen atom, the radius of the \( n^{\text{th}} \) orbit is given by \( r_n = n^2 a_0 \), where \( a_0 = 52.9 \text{ pm} \) is the Bohr radius.

The initial orbit radius is \( 1.3225 \text{ nm} = 1322.5 \text{ pm} \). Hence,

\[ \begin{aligned} n_1^2 &= \frac{1322.5}{52.9} \\ &= 25 \\ n_1 &= 5 \end{aligned} \]

The final orbit radius is \( 211.6 \text{ pm} \). Therefore,

\[ \begin{aligned} n_2^2 &= \frac{211.6}{52.9} \\ &= 4 \\ n_2 &= 2 \end{aligned} \]

Thus, the transition occurs from \( n_1 = 5 \) to \( n_2 = 2 \). The wavelength of the emitted radiation is calculated using the Rydberg relation,

\[ \frac{1}{\lambda} = R\left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right) \]

where \( R = 1.097 \times 10^{7} \text{ m}^{-1} \).

\[ \begin{aligned} \frac{1}{\lambda} &= 1.097 \times 10^{7}\left(\frac{1}{2^2} - \frac{1}{5^2}\right) \\ &= 1.097 \times 10^{7}\left(\frac{1}{4} - \frac{1}{25}\right) \\ &= 1.097 \times 10^{7}\left(\frac{21}{100}\right) \\ &= 2.30 \times 10^{6} \text{ m}^{-1} \end{aligned} \]

Hence,

\[ \begin{aligned} \lambda &= \frac{1}{2.30 \times 10^{6}} \\ &= 4.35 \times 10^{-7} \text{ m} \\ &= 435 \text{ nm} \end{aligned} \]

Therefore, the wavelength of the emitted radiation is approximately \( 435 \text{ nm} \). Since the transition ends at \( n = 2 \), it belongs to the Balmer series, and the wavelength lies in the visible region of the spectrum.

Q57. Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is \(\mathrm{1.6 \times 10^6\ m\ s^{–1}}\), calculate de Broglie wavelength associated with this electron.

Solution

Solution:

According to de Broglie, the wavelength associated with a moving particle is given by \( \lambda = \frac{h}{mv} \), where \( h = 6.626 \times 10^{-34} \text{ J s} \), \( m = 9.11 \times 10^{-31} \text{ kg} \) is the mass of an electron, and \( v = 1.6 \times 10^{6} \text{ m s}^{-1} \) is its velocity.

Substituting the given values,

\[ \begin{aligned} \lambda &= \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(1.6 \times 10^{6})} \\ &= \frac{6.626 \times 10^{-34}}{1.46 \times 10^{-24}} \\ &= 4.54 \times 10^{-10} \text{ m} \end{aligned} \]

Hence, the de Broglie wavelength associated with the electron is \( 4.54 \times 10^{-10} \text{ m} \), which is approximately \( 0.454 \text{ nm} \).

Q58. Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.

Solution

Solution:

For a moving particle, the de Broglie wavelength is given by \( \lambda = \dfrac{h}{mv} \), where \( h = 6.626 \times 10^{-34} \text{ J s} \), \( m \) is the mass of the particle, and \( v \) is its velocity.

The wavelength of neutrons used is \( \lambda = 800 \text{ pm} = 800 \times 10^{-12} \text{ m} = 8.0 \times 10^{-10} \text{ m} \). The mass of a neutron is \( m = 1.675 \times 10^{-27} \text{ kg} \).

Rearranging the de Broglie relation to obtain velocity,

\[ \begin{aligned} v &= \frac{h}{m\lambda} \end{aligned} \]

Substituting the values,

\[ \begin{aligned} v &= \frac{6.626 \times 10^{-34}}{(1.675 \times 10^{-27})(8.0 \times 10^{-10})} \\ &= \frac{6.626 \times 10^{-34}}{1.34 \times 10^{-36}} \\ &= 4.94 \times 10^{2} \text{ m s}^{-1} \end{aligned} \]

Hence, the characteristic velocity of the neutron is approximately \( 4.9 \times 10^{2} \text{ m s}^{-1} \), or about \( 494 \text{ m s}^{-1} \).

Q59. If the velocity of the electron in Bohr’s first orbit is \(\mathrm{2.19 \times 10^6\ ms^{–1}}\), calculate the de Broglie wavelength associated with it.

Solution

Solution:

The de Broglie wavelength of a moving particle is given by \( \lambda = \dfrac{h}{mv} \), where \( h = 6.626 \times 10^{-34} \text{ J s} \), \( m = 9.11 \times 10^{-31} \text{ kg} \) is the mass of the electron, and \( v = 2.19 \times 10^{6} \text{ m s}^{-1} \) is its velocity.

Substituting the given values,

\[ \begin{aligned} \lambda &= \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(2.19 \times 10^{6})} \\ &= \frac{6.626 \times 10^{-34}}{1.995 \times 10^{-24}} \\ &= 3.32 \times 10^{-10} \text{ m} \end{aligned} \]

Hence, the de Broglie wavelength associated with the electron in the first Bohr orbit is \( 3.32 \times 10^{-10} \text{ m} \), which is approximately \( 0.332 \text{ nm} \).

Q60. The velocity associated with a proton moving in a potential difference of 1000 V is \(\mathrm{4.37 \times 10^5\ ms^{–1}}\). If the hockey ball of mass 0.1 kg is moving with this velocity, calcualte the wavelength associated with this velocity.

Solution

Solution:

The de Broglie wavelength associated with any moving particle is given by \( \lambda = \dfrac{h}{mv} \), where \( h = 6.626 \times 10^{-34} \text{ J s} \), \( m \) is the mass of the particle, and \( v \) is its velocity.

For the hockey ball, the mass is \( m = 0.1 \text{ kg} \) and the given velocity is \( v = 4.37 \times 10^{5} \text{ m s}^{-1} \).

Substituting these values into the de Broglie equation,

\[ \begin{aligned} \lambda &= \frac{6.626 \times 10^{-34}}{(0.1)(4.37 \times 10^{5})} \\ &= \frac{6.626 \times 10^{-34}}{4.37 \times 10^{4}} \\ &= 1.52 \times 10^{-38} \text{ m} \end{aligned} \]

Therefore, the wavelength associated with the hockey ball moving at this speed is \( 1.52 \times 10^{-38} \text{ m} \), an extremely small value, showing why wave behaviour is unobservable for macroscopic objects.

Q61. If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is \(\mathrm{h/4πm times 0.05\ nm}\), is there any problem in defining this value.

Solution

Solution:

According to Heisenberg’s uncertainty principle,

\[ \Delta x\,\Delta p \ge \frac{h}{4\pi} \]

The uncertainty in position is given as \( \Delta x = 0.002 \text{ nm} = 2.0 \times 10^{-12} \text{ m} \). Hence, the minimum uncertainty in momentum is

\[ \begin{aligned} \Delta p &= \frac{h}{4\pi \Delta x} \\ &= \frac{6.626 \times 10^{-34}}{4\pi \times 2.0 \times 10^{-12}} \\ &= \frac{6.626 \times 10^{-34}}{2.513 \times 10^{-11}} \\ &= 2.64 \times 10^{-23} \text{ kg m s}^{-1} \end{aligned} \]

Thus, the uncertainty in the momentum of the electron is \( 2.64 \times 10^{-23} \text{ kg m s}^{-1} \).

Now, the momentum of the electron is stated to be

\[ p = \frac{h}{4\pi \times 0.05 \text{ nm}} = \frac{h}{4\pi \times 5.0 \times 10^{-11} \text{ m}} \]

Comparing this with the uncertainty expression, the corresponding positional uncertainty would be

\[ \begin{aligned} \Delta x &= 0.05 \text{ nm} \end{aligned} \]

However, the electron is already localized within \( 0.002 \text{ nm} \), which is much smaller than \( 0.05 \text{ nm} \). This contradicts Heisenberg’s uncertainty principle, because specifying both such a small positional uncertainty and a momentum corresponding to \( 0.05 \text{ nm} \) is not simultaneously possible.

Therefore, there is indeed a problem in defining the given momentum value along with the stated positional accuracy, as it violates the uncertainty principle.

Q62. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists: \[ \begin{array}{c|c} \begin{aligned} (1)&\quad n=4&l=2&&m_l=-2&&m_s=-\frac{1}{2}\\ (2)&\quad n=3&l=2&&m_l=1&&m_s=+\frac{1}{2}\\ (3)&\quad n=4&l=1&&m_l=0&&m_s=+\frac{1}{2}\\ (4)&\quad n=3&l=2&&m_l=-2&&m_s=-\frac{1}{2}\\ (5)&\quad n=3&l=1&&m_l=-1&&m_s=+\frac{1}{2}\\ (6)&\quad n=4&l=1&&m_l=0&&m_s=+\frac{1}{2}\\ \end{aligned} \end{array} \]

Solution

Solution:

For multi-electron atoms, the relative energy of orbitals is determined primarily by the \( n+l \) rule. The orbital with the smaller value of \( n+l \) has lower energy. If two orbitals have the same \( n+l \), the one with smaller \( n \) lies lower in energy. The quantum numbers \( m_l \) and \( m_s \) do not affect energy in the absence of external fields.

First, the values of \( n+l \) are calculated for each set:

\[ \begin{aligned} (1):\; n+l &= 4+2 = 6 \\ (2):\; n+l &= 3+2 = 5 \\ (3):\; n+l &= 4+1 = 5 \\ (4):\; n+l &= 3+2 = 5 \\ (5):\; n+l &= 3+1 = 4 \\ (6):\; n+l &= 4+1 = 5 \end{aligned} \]

The smallest value is for set (5), so this electron has the lowest energy. Next come those with \( n+l = 5 \), namely (2), (3), (4), and (6). Among these, the ones with lower \( n \) have lower energy, so (2) and (4) with \( n=3 \) lie below (3) and (6) with \( n=4 \). Since (2) and (4) have identical \( n \) and \( l \), they possess the same energy. Likewise, (3) and (6) are identical in \( n \) and \( l \), and therefore also have the same energy. Finally, set (1), having the largest value of \( n+l = 6 \), has the highest energy.

Hence, the order of increasing energy is

\[ (5) \;<\; (2) = (4) \;<\; (3) = (6) \;<\; (1) \]

Thus, combinations (2) and (4) are degenerate in energy, combinations (3) and (6) are also degenerate, while (5) is the lowest and (1) is the highest in energy.

Q63. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ?

Solution

Solution:

The effective nuclear charge experienced by an electron depends on two main factors: its distance from the nucleus and the shielding (screening) caused by inner-shell electrons. Electrons present in inner shells shield outer electrons from the full positive charge of the nucleus. As a result, electrons in higher principal quantum number shells feel a smaller effective nuclear charge.

In bromine, the electrons mentioned belong to the \( 2p \), \( 3p \), and \( 4p \) subshells. The \( 2p \) electrons are closest to the nucleus and experience very little shielding, so they feel a large effective nuclear charge. The \( 3p \) electrons are farther away and are shielded by both the K and L shells, so they experience a smaller effective nuclear charge than the \( 2p \) electrons.

The \( 4p \) electrons are the outermost among the given sets. They are shielded by all inner electrons in the K, L, and M shells, and they are also farthest from the nucleus. Consequently, these electrons experience the maximum shielding and therefore the minimum effective nuclear charge.

Hence, the electrons in the \( 4p \) orbital experience the lowest effective nuclear charge.

Q64. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?
(i) 2s and 3s,
(ii) 4d and 4f,
(iii) 3d and 3p.

Solution

Solution:

The effective nuclear charge experienced by an electron depends mainly on its distance from the nucleus and the extent of shielding by inner electrons. Orbitals that are closer to the nucleus and have greater penetrating power experience a larger effective nuclear charge.

(i) Between 2s and 3s orbitals, both have the same shape, but the 2s orbital belongs to a lower principal quantum number and lies much closer to the nucleus. The 3s electron is shielded by all electrons in the \( n=1 \) and \( n=2 \) shells, whereas the 2s electron is shielded only by the 1s electrons. Hence, the 2s orbital experiences a larger effective nuclear charge.

(ii) For 4d and 4f orbitals, both belong to the same principal shell (\( n=4 \)), but their penetration differs. The order of penetration is \( s > p > d > f \). Since 4d electrons penetrate closer to the nucleus than 4f electrons, they experience less shielding and therefore a greater effective nuclear charge than 4f electrons.

(iii) In the case of 3d and 3p orbitals, both have the same principal quantum number, but 3p orbitals have higher penetrating power than 3d orbitals. As a result, 3p electrons can approach the nucleus more closely and are less shielded compared to 3d electrons. Therefore, the 3p orbital experiences the larger effective nuclear charge.

Thus, the orbitals experiencing the larger effective nuclear charge in each pair are 2s, 4d, and 3p respectively.

Q65. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?

Solution

Solution:

Both aluminium and silicon have their unpaired electrons in the \( 3p \) orbital. Since these electrons belong to the same shell and the same type of subshell, the main factor determining the effective nuclear charge is the actual nuclear charge, that is, the number of protons in the nucleus.

Aluminium has atomic number 13, while silicon has atomic number 14. Their inner electronic configurations are very similar, so the shielding provided by inner electrons is nearly the same in both atoms. However, silicon has one extra proton in its nucleus compared to aluminium.

Because of this higher nuclear charge and almost identical shielding, the 3p electron in silicon experiences a stronger attraction toward the nucleus than the 3p electron in aluminium.

Hence, the unpaired 3p electron in silicon experiences more effective nuclear charge than that in aluminium.

Q66. Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Solution

Solution:

The number of unpaired electrons in an atom is obtained from its electronic configuration by applying Hund’s rule, which states that electrons occupy degenerate orbitals singly with parallel spins before pairing begins.

For phosphorus (\( Z = 15 \)), the electronic configuration is \( 1s^2\,2s^2\,2p^6\,3s^2\,3p^3 \). The three electrons in the \( 3p \) subshell occupy three different orbitals, giving

\[ \text{Unpaired electrons in P} = 3 \]

For silicon (\( Z = 14 \)), the configuration is \( 1s^2\,2s^2\,2p^6\,3s^2\,3p^2 \). The two \( 3p \) electrons remain unpaired in separate orbitals, so

\[ \text{Unpaired electrons in Si} = 2 \]

For chromium (\( Z = 24 \)), the stable configuration is \( [Ar]\,3d^5\,4s^1 \). The five \( 3d \) electrons are all unpaired and the single \( 4s \) electron is also unpaired, hence

\[ \text{Unpaired electrons in Cr} = 6 \]

For iron (\( Z = 26 \)), the configuration is \( [Ar]\,3d^6\,4s^2 \). In the \( 3d \) subshell, five electrons occupy separate orbitals and the sixth pairs with one of them, leaving four unpaired electrons. The \( 4s \) electrons are paired. Therefore,

\[ \text{Unpaired electrons in Fe} = 4 \]

For krypton (\( Z = 36 \)), the configuration is \( [Ar]\,3d^{10}\,4s^2\,4p^6 \). All subshells are completely filled, so no unpaired electrons are present:

\[ \text{Unpaired electrons in Kr} = 0 \]

Hence, the numbers of unpaired electrons are: P = 3, Si = 2, Cr = 6, Fe = 4, and Kr = 0.

Q67. (a) How many subshells are associated with n = 4 ?
(b) How many electrons will be present in the subshells having \(\mathrm{m_s}\) value of –1/2 for n = 4 ?

Solution

Solution:

(a) For a given principal quantum number \( n \), the azimuthal quantum number \( l \) can take all integer values from \( 0 \) to \( n-1 \). When \( n = 4 \),

\[ \begin{aligned} l &= 0,\;1,\;2,\;3 \end{aligned} \]

These correspond to the \( 4s \), \( 4p \), \( 4d \), and \( 4f \) subshells. Hence, four subshells are associated with \( n = 4 \).

(b) First, the total number of orbitals present in the \( n = 4 \) shell is

\[ \begin{aligned} \text{Number of orbitals} &= n^2 = 4^2 = 16 \end{aligned} \]

Each orbital can accommodate two electrons with opposite spins. For a specified spin value \( m_s = -\frac{1}{2} \), only one electron per orbital is allowed.

Therefore, the number of electrons having \( m_s = -\frac{1}{2} \) in all subshells of the \( n = 4 \) shell is equal to the total number of orbitals:

\[ \begin{aligned} \text{Electrons with } m_s = -\frac{1}{2} &= 16 \end{aligned} \]

Hence, four subshells are associated with \( n = 4 \), and 16 electrons can have \( m_s = -\frac{1}{2} \) in this shell.