The shortest wavelength of hydrogen atom in Lyman series is . The longest wavelength in Balmer series of is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
The shortest wavelength in Lyman series (H) corresponds to n=∞ → n=1:
1/λ = R·(1²)(1/1² - 0) = R, so R = 1/λ.
The longest wavelength in Balmer series of He⁺ (Z=2) corresponds to n=3 → n=2:
1/λ' = R·Z²·(1/2² - 1/3²) = (1/λ)·4·(1/4 - 1/9) = (1/λ)·4·(5/36) = 5/(9λ).
Therefore λ' = 9λ/5.
1/λ = R·(1²)(1/1² - 0) = R, so R = 1/λ.
The longest wavelength in Balmer series of He⁺ (Z=2) corresponds to n=3 → n=2:
1/λ' = R·Z²·(1/2² - 1/3²) = (1/λ)·4·(1/4 - 1/9) = (1/λ)·4·(5/36) = 5/(9λ).
Therefore λ' = 9λ/5.
[JEE Main 2023 · 29 Jan · Shift 1]