Question
Which of the following statement(s) is (are) correct about the spectrum of hydrogen atom?
JEE Advanced
Physics
Atomic Physics
2021
JEE Advanced 2021 (Paper 1)
JEE Advanced Physics Question (2021) — Solution
Options
- A. The ratio of the longest wavelength to the shortest wavelength in Balmer series is 9 5 .
- B. There is an overlap between the wavelength ranges of Balmer and Paschen series.
- C. The wavelengths of Lyman series are given by 1 + 1 m 2 λ 0 , where λ 0 is the shortest wavelength of Lyman series and m is an integer.
- D. The wavelength ranges of Lyman and Balmer series do not overlap.
Answer
D. The wavelength ranges of Lyman and Balmer series do not overlap.
Step-by-step solution
For Balmer Series: Shortest wavelength, E o 4 = h c λ s Longest wavelength, E o 4 − E 0 9 = h c λ l = 5 E o 36 ∴   λ l λ s = 9 5 Range of energy level in Balmer series, 5 E 0 36 ,   E 0 4 Range of energy level in Paschen series, 7 E 0 144 ,   E 0 9 Hence, no overlap. For Lyman series: For shortest wavelength, E 0 = h c λ 0 For general wavelength, E 0 - E 0 n 2 = h c λ ⇒ 1 1 − 1 n 2 = λ λ 0 ∴   λ = n 2 n 2 − 1 λ 0 or λ = 1 + 1 n 2 − 1 λ 0 = 1 + 1 m 2 λ 0 as m should be an integer. Range of energy level in Lyman series, 3 E 0 4 ,   E 0 ⇒ Lyman and Balmer series wavelengths do not overlap.
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