An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let $\lambda_n$, $\lambda_g$ be the de Broglie wavelength of the electron in the $n^{\text{th}}$ state and the ground state respectively. Let $\Lambda_n$ be the wavelength of the emitted photon in the transition from the $n^{\text{th}}$ state to the ground state. For large n, (A, B are constants)

(B) $\Lambda_n \approx A+\frac{B}{\lambda_n^2}$

(A) $\Lambda_n^2 \approx \lambda$

(C) $\Lambda_n \approx A+B\lambda_n$

(D) $\Lambda_n^2 \approx A+B\lambda_n^2$

JEE Main · 2018 · 08 Apr 2018 · Offline · Modern Physics

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let

λ_{n}

λ_{g}

be the de Broglie wavelength of the electron in the

n^{th}

state and the ground state respectively. Let

Λ_{n}

be the wavelength of the emitted photon in the transition from the

n^{th}

state to the ground state. For large n, (A, B are constants)

Answer: (B)

Bohr quantisation

2 π r = n λ_{n}

makes the electron's de Broglie wavelength grow with the orbit,

λ_{n} \propto n

, so

n^{2} \propto λ_{n}^{2}

. The photon emitted in the transition to the ground state has

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

, which for large

n

gives

Λ_{n} = \frac{1}{R} (1 - \frac{1}{n ^{2}})^{- 1} \approx \frac{1}{R} (1 + \frac{1}{n ^{2}})

. Replacing

\frac{1}{n ^{2}}

by a constant times

\frac{1}{λ _{n}^{2}}

, this takes the form

Λ_{n} \approx A + \frac{B}{λ _{n}^{2}}

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option