# NCERT Solution : Atoms

Q:1. Choose the correct alternative from the clues given at the end of each statement:

(a) The size of the atom in Thomson’s model is….. the atomic size in Rutherford’s model (much greater than/no different from/much less than)

(b) In the ground state of………, electrons are in stable equilibrium, while in …… electrons always experience a net force (Thomson’s model/Rutherford’s model).

(c) A classical atom based on………is doomed to collapse (Thomson’s model/ Rutherford’s model).

(d) An atom has a nearly continuous mass distribution in ….. but has a highly non uniform mass distribution in ….(Thomson’s model/Rutherford’s model).

(e) The positively charged part of the atom possesses most of the mass of the atom in …..(Rutherford’s model/ both the models).

Soln:
(a) No different from

(b) Thomson’s model, Rutherford’s model

(c) Rutherford’s model

(d) Thomson’s model, Rutherford’s model

(e) Both the models.

Q:2. Suppose you are given a chance to repeat the α-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperature below 14 K). What result do you expect ?

Sol. In the α-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen is less the mass of incident α-particles (6.64 × 10-27 kg). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α-particles would not bounce back, if solid hydrogen is used in the α-particles scattering experiment.

Q:3. What is the shortest wavelength present in the Paschen series of spectral lines ?
(R = 1.097 × 107 m-1)

Sol. Using formula for Paschen series.

$\large \frac{1}{\lambda} = R [\frac{1}{3^2} – \frac{1}{n_2^2} ]$ ;  n2 = 4, 5, 6,…

For shortest wavelength, n2 = ∞

$\large \frac{1}{\lambda} = R [\frac{1}{3^2} – \frac{1}{\infty} ] = \frac{R}{9}$

λ = 9/R = 820.4 nm

Q:4. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level ?

Sol. Separation of two energy level in an atom.

ΔE = 2.3 eV = 2.3 × 1.6 × 10-19 = 3.68 × 10-19 J

Let ν be the frequency of radiation emitted, when the atom transition from the upper level to the lower level.

Energy of radiation , ΔE = h ν

$\large \nu = \frac{\Delta E}{h} = \frac{3.68 \times 10^{-19}}{6.63 \times 10^{-34}}$

= 5.6 × 1015 Hz

Q: 5. The ground state energy of H-atom is -13.6 eV. What are the kinetic and potential energies of electron in this state ?

Sol: Total Energy , E = -13.6 eV

K.E = -E = 13.6 eV

P.E = -2KE = -27.2 eV

Q:6. A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

Sol: Energy of photon absorbed ,

$\large E = E_2 – E_1$

$\large E = 13.6 (\frac{1}{n_1^2} – \frac{1}{n_2^2} ) eV$

$\large E = 13.6 (\frac{1}{1^2} – \frac{1}{4^2} ) eV$

$\large = 13.6 \times \frac{15}{16} eV$

$\large = 13.6 \times \frac{15}{16} \times 1.6 \times 10^{-19} J$

= 2.04  × 10-18 J

$\large \lambda = \frac{h c}{E} = \frac{6.6 \times 10^{-34}\times 3 \times 10^8}{2.04 \times 10^{-18}}$

= 9.7 × 10-8 m

= 97 nm

Frequency , $\large \nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{9.7 \times 10^{-8}}$

= 3.1 × 1015Hz

Q:7. (i) Using the Bohr’s model, calculate the speed of the electron in a H- atom in the n = 1 , 2 and 3 levels.

(ii) Calculate the orbital period in each of these levels.

Sol: (a) Using , $\large v = \frac{c}{n} \times \alpha$ ; $Where \; \alpha = \frac{2\pi K e^2}{c h} = 0.0073$

$\large v_1 = \frac{3 \times 10^8}{1} \times 0.0073$

= 2.19 × 106 m/s

$\large v_2 = \frac{3 \times 10^8}{2} \times 0.0073$

= 1.095 × 106 m/s

$\large v_3 = \frac{3 \times 10^8}{3} \times 0.0073$

= 7.3 × 105 m/s

(b) Orbital Speed $\large T = \frac{2\pi r}{v}$

As , r1 = 0.53 × 10-10 m

$\large T_1 = \frac{2 \pi \times 0.53 \times 10^{-10}}{2.19 \times 10^6}$

= 1.52 × 10-16 sec

r2 = 4 r1 and v2 = v1/2

T2 = 8 T1

= 1.216 × 10-15 sec

r3 = 9 r1 and v3 = v1/3

T3 = 27 T1

= 4.1 × 10-15 sec

Q:8. The radius of the innermost electron orbit of a H-atom is 5.3 × 10-11 m. What are the radii of the n=2 and n=3 orbits ?

Sol: $\large r \propto n^2$

r2 = 4 r1

r2 = 4 × 5.3 × 10-11 m

= 2.12 × 10-10 m

r3 = 9 r1

r3 = 9 × 5.3 × 10-11 m

= 4.77 × 10-10 m

Q: 9. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Sol: In ground state, energy of gaseous hydrogen at room temperature = -13.6 eV. When it is bombarded with 12.5 eV electron beam, the energy becomes -13.6 + 12.5 = -1.1 eV. The electron would jump from n = 1 to n = 3, where E3 = -13.6/32 = – 1.5 eV. On de-excitation the electron may jump from n = 3 to n = 2 giving rise to Balmer series. It may also jump from n = 3 to n = 1, giving rise to Lyman series.

Q:10. In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth = 6.0 × 1024 kg)

Soln:
Here, r = 1.5 × 1011 m, ? = 3 × 104 m/s, m = 6.0 × 1024 kg

According to Bohr’s model, $\large m v r = \frac{n h}{2 \pi}$

$\large n = \frac{2 \pi m v r}{h}$

$\large n = \frac{2 \times 3.14 \times 6 \times 10^{24} \times 3 \times 10^4 \times 1.5 \times 10^{11} }{6.6 \times 10^{-34}}$

n = 2.57 × 1074, which is too large.

Q:11. Answer the following: (Which help you understand the difference between Thomson’s model and Rutherford’s model better).
(a) In the average angle of deflection of  particles by a thin gold foil predicted by Thomson’s model much less, about the same or much greater than that predicted by Rutherford’s atom model?
(b) Is the probability of backward scattering (i.e. scattering of  particles at angle greater than 90°) predicted by Thomson’s model, much less, about the same, or much greater than that predicted by Rutherford’s model?
(c) Keeping other factors fixed, it is found experimentally, that for a small thickness t, the number of alpha particles scattered at moderate angles is proportional to t. What clues does this linear dependence on t provide?
(d) In which atom model, is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of alpha particles by a thin foil?

Solution:
(a) About the same. This is because we are talking of average angle of deflection.

(b) Much less, because in Thomson’s model, there is no such massive central core called the nucleus as in Rutherford’s model.

(c) This suggests that scattering is predominantly due to a single collision, because change of a single collision increases with the number of target atoms which increases linearly with the thickness of the foil.

(d) In Thomson model, positive charge is uniformly distributed in the spherical atom. Therefore, a single collision causes very little deflection. Therefore, average scattering angle can be explained only by considering multiple scattering. Thus it is wrong to ignore multiple scattering in Thomson’s model.

On the contrary, in Rutherford’s model, most of the scattering comes from a single collision. Therefore, multiple scattering may be ignored as a first approximation.

Q:12. The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10-40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

Solution:

The radius of the first Bohr orbit of a hydrogen atom is $\displaystyle r_0 = \frac{4\pi \epsilon_0(h/2\pi)^2}{m_e e^2}$
If we consider the atom bound by the gravitational force $\displaystyle (=\frac{G m_p m_e}{r^2} )$ , we should replace $\displaystyle (=\frac{e^2}{4\pi \epsilon_0} )$ by $\displaystyle (G m_p m_e )$.

In that case, radius of first Bohr orbit of hydrogen atom would be given by $\displaystyle r_0 = \frac{(h/2\pi)^2}{G m_p m_e^2}$

Putting the standard values, we get $\displaystyle r_0 = \frac{6.6 \times 10^{-34}/2\pi}{6.67 \times 10^{-11}\times 1.67 \times10^{-27} \times (9.1 \times10^{-31})^2}$ = 1.2 × 1029 metre

This is much greater than the estimated size of the whole universe!

Q:15. The total energy of electron in the first excited state of hydrogen atom is about -3.4 eV
(a) What is kinetic energy of electron in this state?
(b) What is potential energy of electron in this state?
(c) Which of the answers above would change if the choice of zero of potential energy is changed ?

Solution:
We know Kinetic energy of electron $\displaystyle = \frac{KZe^2}{2r}$ and P.E. of electron $\displaystyle = – \frac{KZe^2}{r}$

P.E. = -2 (Kinetic energy)

In this calculation, electric potential and hence potential energy is zero at infinity.

Total energy = P.E. + K.E. = -2 KE + KE = -KE

(a) In the first excited state, total energy = -3.4 eV

KE = -(-3.4eV) = + 3.4 eV

(b) P.E. of electron in this first excited state = -2KE = -2 × 3.4 = – 6.8 eV

(c) If zero of potential energy is changed, K.E. does not change and continues to be +3.4 eV. However, the P.E. and total energy of the state would change with the choice of zero of potential energy.

Q:16. If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?

Solution: Bohr’s quantization postulate is in terms of Planck’s constant (h). But angular momenta associated with planetary motion are ≈ 1070 h (for earth). In terms of Bohr’s quantisation postulate, this will correspond to n ≈ 1070. For such large values of n, the differences in successive energies and angular momenta of the quantised levels are so small, that the levels can be considered as continuous and not discrete.

Q:17. Obtain the first Bohr radius and the ground state energy of muonic hydrogen atom (i.e. an atom in which a negatively charged muon (μ) of mass about 207 me revolves around a proton).

Solution: A muonic hydrogen is the atom in which a negatively charged muon of mass about 207 me revolves around a proton.

In Bohr’s atom model, as, r ∝ 1/m

rμ/re =me/mμ =me/(207me )= 1/207

Here re is radius of first orbit of electron in hydrogen atom = 0.50 A° = 0.53 × 10-10 m

rμ = re/207 = (0.53×10^(-10))/207 = 2.56×10^(-13) m

Again, in Bohr’s atom model.

E ∝ m

Eμ/Ee = mμ/me = 207me/me ,Eμ = 207Ee

As ground state energy of electron in hydrogen atom is

Ee = -13.6 eV

Eμ = 207 (-13.6)eV = – 2815.2eV = -2.8152 keV