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.