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

(B) $\sqrt{\frac{\lambda R}{\lambda R-1}}$

(A) $\sqrt{\frac{\lambda R}{\lambda-1}}$

(C) $\sqrt{\frac{\lambda}{\lambda R-1}}$

(D) $\sqrt{\frac{\lambda R^{2}}{\lambda R-1}}$

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Modern Physics

Hydrogen atom from excited state comes to the ground by emitting a photon of wavelength

λ

. The value of principal quantum number 'n' of the excited state will be:
(R: Rydberg constant)

Answer: (B)

Energy of the photon emitted in transition from level

n

to ground state (

n = 1

E = E_{n} - E_{1} = \frac{- R h c}{n ^{2}} - \frac{- R h c}{1 ^{2}} = R h c (1 - \frac{1}{n ^{2}})

Also,

E = \frac{h c}{λ}

. Equating:

\frac{h c}{λ} = R h c (1 - \frac{1}{n ^{2}})

\frac{1}{λ} = R (1 - \frac{1}{n ^{2}})

\frac{1}{λ R} = 1 - \frac{1}{n ^{2}}

\frac{1}{n ^{2}} = 1 - \frac{1}{λ R} = \frac{λ R - 1}{λ R}

n^{2} = \frac{λ R}{λ R - 1}

n = \frac{λ R}{λ R - 1}

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option