For any given series of spectral lines of atomic hydrogen, let be the difference in maximum and minimum frequencies in cm. The ratio is
- A.27 : 5
- B.5 : 4
- C.9 : 4
- D.4 : 1
Show answer and solution
Correct option: C
Solution
For the Lyman series, transitions end at n=1. The max frequency is n=∞→1: ν̄_max = R(1/1² - 0) = R, and min frequency is n=2→1: ν̄_min = R(1/1² - 1/2²) = 3R/4. So Δν̄_Lyman = R - 3R/4 = R/4.
For the Balmer series, transitions end at n=2. The max frequency is n=∞→2: ν̄_max = R/4, and min frequency is n=3→2: ν̄_min = R(1/4 - 1/9) = 5R/36. So Δν̄_Balmer = R/4 - 5R/36 = 9R/36 - 5R/36 = R/9.
Therefore Δν̄_Lyman / Δν̄_Balmer = (R/4)/(R/9) = 9/4, giving the ratio 9:4.
For the Balmer series, transitions end at n=2. The max frequency is n=∞→2: ν̄_max = R/4, and min frequency is n=3→2: ν̄_min = R(1/4 - 1/9) = 5R/36. So Δν̄_Balmer = R/4 - 5R/36 = 9R/36 - 5R/36 = R/9.
Therefore Δν̄_Lyman / Δν̄_Balmer = (R/4)/(R/9) = 9/4, giving the ratio 9:4.
[JEE Main 2019 · 9 Apr · Shift I]