Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength λ. The value of principal quantum number ' n ' of the excited state will be: ( R: Rydberg constant) (a) √((λ R)/(λ - 1)) (b) √((λ R)/(λ R - 1)) (c) √((λ)/(λ R - 1)) (d) √λ R²λ R - 1

Answer: (b). (1)/(λ) = R(1 - 1/n²) 1/n² = 1 - (1)/(λ R) = (λ R - 1)/(λ R) n = √((λ R)/(λ R - 1)), so the answer is (b).

Hydrogen atom from excited state comes to the ground state by… | JEE Main 2022 (25 Jul, Shift 2)

Q40JEE Main 2022 · 25 JUL · SHIFT 2Medium

Transitions, Photon Emission & Excitation

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength $\lambda$ . The value of principal quantum number ' $n$ ' of the excited state will be: ( $R:$ Rydberg constant)

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Answer: (b)

$\frac{1}{\lambda} = R\left(1 - \frac{1}{n^2}\right)$
$\frac{1}{n^2} = 1 - \frac{1}{\lambda R} = \frac{\lambda R - 1}{\lambda R}$
$n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$ , so the answer is (b).

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