An electron in a hydrogen atom falls from n = 4 to n = 2. The wavelength of the emitted photon (R_H = 1.097 × 10⁷ m⁻¹) is closest to:
A656 nm
B486 nm
C434 nm
D410 nm
Answer & Solution
Correct answer: B. 486 nm
1. 1/λ = R_H (1/n₁² − 1/n₂²) with n₁ = 2, n₂ = 4.
2. = 1.097 × 10⁷ × (1/4 − 1/16) = 1.097 × 10⁷ × 3/16 = 2.057 × 10⁶ m⁻¹.
3. λ = 1 / 2.057 × 10⁶ ≈ 4.86 × 10⁻⁷ m = 486 nm.
4. This is Hβ of the Balmer series — blue-green line.
_Source: NCERT Class 11 Chem Ch 2 §2.3 Hydrogen Spectrum (Balmer)_
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