Chapter 12: Atoms
Bohr's Model:
• En = −13.6/n² eV (for hydrogen)
• rn = 0.529 n² Å
• ν = 13.6(1/n₁² − 1/n₂²) Hz
• Angular momentum: L = nh/(2π)
Q1. Find the radius of the second orbit and the energy of an electron in the second orbit of hydrogen atom.
Solution
r₂ = r₀ × n² = 0.529 × 4 = 2.116 Å
E₂ = −13.6/n² = −13.6/4 = −3.4 eV
Q2. Calculate the wavelength of the spectral line corresponding to the transition from n = 4 to n = 2 in hydrogen atom.
Solution
1/λ = R(1/2² − 1/4²) = R(1/4 − 1/16) = R(3/16)
λ = 16/(3R) = 16/(3 × 1.097 × 107)
= 16/(3.291 × 107)
= 4.86 × 10−7 m = 486 nm (H-beta line, visible)
Q3. Find the ionization energy of hydrogen atom in the ground state.
Solution
Ionization energy = −E₁ = −(−13.6 eV)
= 13.6 eV
Q4. What are the limitations of Bohr's model of the atom?
Solution
1. It only works for hydrogen-like atoms (single electron systems).
2. It cannot explain the fine structure of spectral lines.
3. It violates Heisenberg's uncertainty principle.
4. It doesn't explain the intensity of spectral lines.
5. It treats electrons as particles in definite orbits, which contradicts the wave nature of electrons.
Q5. The total energy of an electron in the third orbit of hydrogen atom is −1.51 eV. Find its kinetic energy and potential energy.
Solution
For hydrogen atom:
K.E. = −E = 1.51 eV
P.E. = 2E = 2(−1.51) = −3.02 eV
Total E = K.E. + P.E. = 1.51 − 3.02 = −1.51 eV ✓
Q6. Find the wavelength of the first line of the Balmer series of hydrogen atom.
Solution
Balmer series: 1/λ = R(1/2² − 1/n²), n = 3, 4, 5...
First line (n = 3):
1/λ = R(1/4 − 1/9) = R(5/36)
λ = 36/(5R) = 36/(5 × 1.097 × 107)
= 36/(5.485 × 107)
= 6.56 × 10−7 m = 656 nm (Hα line, red)
Q7. Using Bohr's model, find the orbital speed and the time period of revolution of an electron in the first orbit of hydrogen atom.
Solution
v = e²/(2ε₀nh) = e²/(2ε₀h) for n=1
= (1.6 × 10−19)²/2(8.85 × 10−12)(6.63 × 10−34)
= 2.56 × 10−38/(1.173 × 10−44)
= 2.18 × 106 m/s
T = 2πr₁/v = 2π(0.529 × 10−10)/(2.18 × 106)
= 1.52 × 10−16 s
Q8. A hydrogen atom in the ground state absorbs a photon of energy 12.75 eV. Find the excited state it reaches. What are the possible wavelengths of the emitted photons?
Solution
E₁ = −13.6 eV
Energy after absorption: −13.6 + 12.75 = −0.85 eV
En = −13.6/n² = −0.85
n² = 13.6/0.85 = 16
n = 4 (4th excited state)
Possible transitions: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1
Six possible wavelengths (corresponding to Lyman, Balmer, and Paschen series lines).
Q9. The energy levels of a certain atom are given as: E₁ = −10 eV, E₂ = −5 eV, E₃ = −2 eV. Find the ionization energy and the wavelength of radiation emitted when the atom transitions from E₃ to E₁.
Solution
Ionization energy = −E₁ = 10 eV
For E₃ → E₁ transition:
ΔE = E₃ − E₁ = −2 − (−10) = 8 eV
λ = hc/E = 12400/8 = 1550 Å
Q10. Explain Rutherford's α-scattering experiment. What conclusions did Rutherford draw from the observations?
Solution
Experiment: α-particles (He² nuclei) were bombarded on a thin gold foil.
Observations:
1. Most α-particles passed straight through
2. Some were deflected through small angles
3. A very few (1 in 8000) rebounded back (>90°)
Conclusions:
1. Most of the atom is empty space
2. Positive charge is concentrated in a tiny nucleus
3. Nucleus size ≈ 10−15 m (atom size ≈ 10−10 m)
4. Almost all mass is in the nucleus
Q11. Find the kinetic energy, potential energy, and total energy of an electron in the second orbit of hydrogen atom.
Solution
E₂ = −13.6/4 = −3.4 eV
K.E. = −E = 3.4 eV
P.E. = 2E = −6.8 eV
Total E = K.E. + P.E. = −3.4 eV
Q12. The wavelength of the second line of the Lyman series is 1026 Å. Find the wavelength of the first line of this series.
Solution
Second line of Lyman: n = 3 → n = 1
1/1026 = R(1/1 − 1/9) = R(8/9)
R = 9/(8 × 1026) = 1.0965 × 107 m−1
First line (n = 2 → n = 1):
1/λ = R(1 − 1/4) = R(3/4)
λ = 4/(3R) = 4/(3 × 1.0965 × 107)
= 1215 Å (Lymanα line)
Q13. A hydrogen atom makes a transition from n = 4 to n = 2. Calculate the frequency of the emitted radiation. (R = 1.097 × 107 m−1)
Solution
1/λ = R(1/2² − 1/4²) = R(1/4 − 1/16) = R(3/16)
f = c/λ = cR(3/16)
= (3 × 108)(1.097 × 107)(3/16)
= 3 × 1.097 × 3/16 × 1015
= 6.17 × 1014 Hz (Hβ line, blue-green)