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.

Correct answer: B. ∧ _ n A + B λ _ n ^ 2 2 = n ⋋ _ n ⋋ _ n = (2 )/(n) ∴ r = r _ o n ^ 2 ⋋ _ n = (2π)/(n) r _ o n ^ 2 = n(2π r _ o ) . . (i) 1 ∧ _ n = R\ 1 1 ^ 2 - 1 n ^ 2 \ ∧ _ n = (1)/(R)\ 1 + 1 n ^ 2 - 1 \ ∧ _ n = (1)/(R)\ 1 + 1 n ^ 2 \ . . (ii) from ( i ) and ( ii ) ∧ _ n = (1)/(R)\ 1 + 4 n ^ 2 r _ o ^ 2 ⋋ _ n ^ 2 \ = A + B ⋋ _ n ^ 2

JEE Main 20188 April

Exam: JEE Main
Year: 2018
Date: 8 April
Subject: Physics
Chapter: Atomic Physics

JEE Main 2018 · 8 April · Physics · Atomic Physics

An electron from various excited states of hydrogen atom emit...

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}^{th}

state and the ground state respectively. Let

{∧}_{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)

Options

A ${∧}_{n}^{2} \approx \lambda$
B ${∧}_{n} \approx A + \dfrac{B}{{\lambda}_{n}^{2}}$
C ${∧}_{n} \approx A + B{\lambda}_{n}$
D ${∧}_{n}^{2} \approx A + B{\lambda}_{n}^{2}$

Correct answer

B. ${∧}_{n} \approx A + \dfrac{B}{{\lambda}_{n}^{2}}$

Step-by-step solution

2\pir = n{⋋}_{n}

{⋋}_{n} = \dfrac{2\pir}{n}

∴ r = {r}_{o}{n}^{2}

{⋋}_{n} = \dfrac{2\pi}{n}{r}_{o}{n}^{2} = n\left(2\pi{r}_{o}\right) \ldots . . \left(i\right)

\dfrac{1}{{∧}_{n}} = R\left\{\dfrac{1}{{1}^{2}} - \dfrac{1}{{n}^{2}}\right\}

{∧}_{n} = \dfrac{1}{R}\left\{1 + \dfrac{1}{{n}^{2} - 1}\right\}

{∧}_{n} = \dfrac{1}{R}\left\{1 + \dfrac{1}{{n}^{2}}\right\} \ldots \ldots \ldots . . \left(ii\right)

from

( i )

and

( ii )

{∧}_{n} = \dfrac{1}{R}\left\{1 + \dfrac{4{n}^{2}{r}_{o}^{2}}{{⋋}_{n}^{2}}\right\} = A + \dfrac{B}{{⋋}_{n}^{2}}

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