NCERT Class 12 Physics Chapter 12 Atoms

NCERT Class 12 Physics Chapter 12 Atoms Solutions, NCERT Solutions For Class 12 Physics to each chapter is provided in the list so that you can easily browse throughout different chapters NCERT Class 12 Physics Chapter 12 Atoms Question Answer and select needs one.

NCERT Class 12 Physics Chapter 12 Atoms

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Also, you can read the CBSE book online in these sections NCERT Class 12 Physics Chapter 12 Atoms Solutions by Expert Teachers as per NCERT (CBSE) Book guidelines. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 12 Physics Chapter 12 Atoms Solutions for All Subjects, You can practice these here.

Atoms

Chapter: 12

EXERCISE

1. Choose the correct alternative from the clues given at the end of the 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.)

Ans: Not different from.  

(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.) 

Ans: Thomson’s model; Rutherford’s model.

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

Ans: Rutherford’s model.

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

Ans: Thomson’s model, Rutherford’s model.

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

Ans: Both the models. 

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

Ans: In Rutherford’s alpha-particle scattering experiment, if we use a thin sheet of solid hydrogen instead of gold foil, we wouldn’t observe significant large-angle scattering. This is because hydrogen nuclei (protons) are much smaller and lighter than gold nuclei. When an alpha particle collides with a proton, the proton’s small mass prevents it from deflecting the alpha particle significantly. It’s like a football hitting a tennis ball: the football’s momentum is so much greater that it barely affects the tennis ball’s direction. So, in the case of hydrogen, most alpha particles would pass straight through the sheet without being deflected much.

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

Ans: Given:

E = 2.3 eV

h = 4.1357 × 10^-15 eV

E = h𝑓

Rearrange the formula:

𝑓 = E/h

= 2.3 eV / (4.1357 × 10^-15 eV)

= 5.56 × 10¹⁴ Hz. 

4. The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Ans: Given:

E = – 13.6 eV

K = – (- 13.6 eV) = 13.6 eV

U = 2E

= 2 × (-13.6 eV)

= – 27.2 eV. 

5. 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.

Ans: Given:

E_n = – 13.6 eV × 1/n²

E₁ = – 13.6 eV

E₄ = – 13.6 eV × 1/42 

= – 0.85 eV

ΔE = E₄ – E₁ 

= – 0.85 eV – (- 13.6 eV )

= 12.75 eV.

Wavelength of the photon:

E = hc / λ

h = 4.1357 × 10^-15 eV

c = 3 × 10⁸ m/s

λ = hc/ ΔE

= (4.1357 × 10^-15) × ( 3 × 10⁸ m/s) / 12.75 eV

= 97.5 nm

Frequency of the photon:

E = hf

f = E/h

= 12.75 eV / 4.1357 × 10^-15 

= 3.08 × 10¹⁵ Hz. 

6. (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels.

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

Ans: 

= 1.5 × 10^-16 s

T₂ = T₁ × (2)3 = 1.52 × 10^-16 × 8 s

= 1.22 × 10^-15 s

T₃ = T₁ × (3)3 = 1.52 × 10^-16 × 27 s

= 4.11 × 10^-15 s. 

7. The radius of the innermost electron orbit of a hydrogen atom is 5.3 ×10^–11 m. What are the radii of the n = 2 and n =3 orbits?

Ans: Given:

r₁ = 5.3 × 10^-11 m

rn = n2 × r₁

Calculate:

For n² = 2

r₂ = 2² × r₁ 

= 4 × 5.3 × 10^-11 m

= 2.12 × 10^-10 m

For n = 3:

r₃ = 3² × r₁ 

= 9 × 5.3 × 10^-11 m

= 4.77 × 10^-10 m. 

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

Ans: 

When a 12.5 eV electron beam bombards hydrogen gas, the atoms can be excited up to the n = 3 level. However, the electrons can also be excited to the n = 2 level, as the energy difference between n = 1 and n = 2 is 10.2 eV. They can then transition to lower energy levels, emitting photons. The emitted wavelengths will include lines from the Balmer series and the Lyman series Thus, the emitted wavelengths cover both visible and ultraviolet regions.

9. 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.) 

Ans: Given:

m = 6.0 × 10²⁴ kg

v = 3.0 × 10⁴ m/s

r = 1.5 × 10¹¹

L = mvr

= (6.0 × 10²⁴ kg) × (3.0 × 10⁴ m/s) × (1.5 × 10¹¹ m)

= 2.7 × 10⁴⁰ kg m²/s