Case Study Questions for Class 12 Physics Chapter 12 Atoms
Case Study Question 1:
The spectral series of hydrogen atom were accounted for by Bohr using the relation
where, R=Rydberg constant = 1.097 x 107 m-1
Lyman series is obtained when an electron jumps to first orbit from any subsequent orbit. Similarly, Balmer series is obtained when an electron jumps to 2nd orbit from any subsequent orbit. Paschen series is obtained when an electron jumps to 3rd orbit from any subsequent orbit. Whereas Lyman series in U.V. region, Balmer series is in visible region and Paschen series lies in infrared region. Series limit is obtained when n2=∞.
(i) The wavelength of first spectral line of Lyman series is
(a) 1215.4 A0
(b) 1215.4 cm
(c) 1215.4 m
(d) 1215. 4 mm
(ii) The wavelength limit of Lyman series is
(a) 1215.4 A0
(b) 511.9 A0
(c) 951.6 A0
(d) 911.6 A0
(iii) The frequency of first spectral line of Balmer series is
(a) 1.097 x 107 Hz
(b) 4.57 x 1014 Hz
(c) 4.57 x 1015 Hz
(d) 4.57 x 1016 Hz
(iv) Which of the following transitions in hydrogen atom emit photon of highest frequency?
(a) n=1 to n=2
(b) n=2 to n=6
(c) n=6 to n=2
(d) n=2 to n=1
(v) The ratio of minimum to maximum wavelength in Balmer series is
(a) 5 : 9
(b) 5 : 36
(c) 1 : 4
(d) 3 : 4
Answers and Explanations:
(i) Correct Option: (a) 1215.4 Å
Explanation:
For the first Lyman line, the transition is: n₂ = 2 → n₁ = 1
Using the formula:
1/λ = R (1/1² − 1/2²)
= R (1 − 1/4) = (3/4)R
λ = 4 / (3R)
Substituting R = 1.097 × 10⁷ m⁻¹ gives:
λ ≈ 121.54 nm = 1215.4 Å.
(ii) Correct Option: (d) 911.6 Å
Explanation:
Wavelength limit (shortest wavelength) occurs when n₂ → ∞ for Lyman series:
1/λlimit = R (1/1² − 1/∞²) = R
λlimit = 1/R = 1 / (1.097 × 10⁷)
λlimit ≈ 9.116 × 10⁻⁸ m = 911.6 Å.
(iii) Correct Option: (b) 4.57 × 10¹⁴ Hz
Explanation:
First Balmer line: n₂ = 3 → n₁ = 2
Using:
1/λ = R (1/2² − 1/3²)
= R (1/4 − 1/9) = R (5/36)
λ = 36 / (5R)
Substitute R = 1.097 × 10⁷ m⁻¹ to get λ ≈ 656 nm.
Frequency: ν = c / λ = (3×10⁸) / (6.56×10⁻⁷)
ν ≈ 4.57 × 10¹⁴ Hz.
(iv) Correct Option: (d) n = 2 → n = 1
Explanation:
Higher frequency corresponds to higher energy difference ΔE.
Energy difference increases as the transition ends in a lower orbit.
Among options, the largest energy drop is:
n = 2 → 1 (Lyman series transition).
Hence maximum frequency photon is emitted.
(v) Correct Option: (b) 5 : 36
Explanation:
Balmer series: n₁ = 2
Maximum wavelength (longest) occurs for the first line: n₂ = 3 → 2
1/λmax = R (1/2² − 1/3²) = R (5/36)
So λmax ∝ 36/5.
Minimum wavelength (shortest) occurs at the series limit: n₂ → ∞ → 2
1/λmin = R (1/2² − 0) = R/4
So λmin ∝ 4.
Ratio (minimum : maximum):
λmin : λmax = 4 : (36/5) = 5 : 36.
