If an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength λ . When it jumps from the fourth orbit to the third orbit, the corresponding wavelength of the photon

Correct answer: C. (20)/(7)λ (1)/(λ) = R[1 n _ 1 ^ 2 - 1 n _ 2 ^ 2 ] ⇒ λ = 1 R( 1 2 ^ 2 - 1 3 ^ 2 ) ⇒ λ ^ ′ = 1 R( 1 3 ^ 2 - 1 4 ^ 2 ) ⇒ λ ^ ′ λ = ( 1 2 ^ 2 - 1 3 ^ 2 ) ( 1 3 ^ 2 - 1 4 ^ 2 ) ⇒ λ ^ ′ = (20)/(7)λ .

NEET 2016

Exam: NEET
Year: 2016
Subject: Physics
Chapter: Atomic Physics

NEET 2016 · Physics · Atomic Physics

If an electron in a hydrogen atom jumps from the third orbit to the...

If an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength

\lambda

. When it jumps from the fourth orbit to the third orbit, the corresponding wavelength of the photon will be

Options

A $\dfrac{16}{25}\lambda$
B $\dfrac{9}{16}\lambda$
C $\dfrac{20}{7}\lambda$
D $\dfrac{20}{13}\lambda$

Correct answer

C. $\dfrac{20}{7}\lambda$

Step-by-step solution

$\dfrac{1}{\lambda} = R\left[1{n}_{1}^{2} - 1{n}_{2}^{2}\right]$

$⇒ \lambda = \dfrac{1}{R\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{3}^{2}}\right)}$

$⇒ {\lambda}^{′} = \dfrac{1}{R\left(\dfrac{1}{{3}^{2}} - \dfrac{1}{{4}^{2}}\right)}$

$⇒ \dfrac{{\lambda}^{′}}{\lambda} = \dfrac{\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{3}^{2}}\right)}{\left(\dfrac{1}{{3}^{2}} - \dfrac{1}{{4}^{2}}\right)} ⇒ {\lambda}^{′} = \dfrac{20}{7}\lambda$ .

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