When an electron in hydrogen atom jumps from the third orbit to the second orbit it emits a photon of wavelength lambda?

Option 3 : \(\frac{{20}}{7}\lambda \)

Free

15 Questions 15 Marks 12 Mins

 When electron jumps from higher orbit to lower orbit then, wavelength of emitted photon is given by,

\(\frac{1}{\lambda } = R\left( {\frac{1}{{n_f^2}} - \frac{1}{{n_i^2}}} \right)\) 

So, \(\frac{1}{\lambda } = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) = \frac{{5R}}{{36}}\) 

and \(\frac{1}{{\lambda '}} = R\left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) = \frac{{7R}}{{144}}\) 

∴ \(\lambda ' = \frac{{144}}{7} \times \frac{{5\lambda }}{{36}} = \frac{{20\lambda }}{7}\)

Ace your Atoms preparations for The Line Spectra of the Hydrogen Atom with us and master Physics for your exams. Learn today!

India’s #1 Learning Platform

Start Complete Exam Preparation

Daily Live MasterClasses

Practice Question Bank

Mock Tests & Quizzes

Get Started for Free Download App

Trusted by 3.3 Crore+ Students

Tardigrade - CET NEET JEE Exam App

© 2022 Tardigrade®. All rights reserved

When 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 third orbit, the wavelength emitted by the photon will be ______.

• `20/7lambda`

• `16/25lambda`

• `20/13lambda`

• `9/16lambda`

When 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 third orbit, the wavelength emitted by the photon will be `underline(20/7lambda)`.

Explanation:

`1/lambda=R(1/2^2-1/3^2)=R(1/4-1/9)=(Rxx5)/36`

`1/lambda^'=R(1/3^2-1/4^2)=R(1/9-1/16)=Rxx7/144`

`lambda^'/lambda=5/36xx144/7=20/7`

`thereforelambda^'=20/7lambda`

Concept: Bohr’s Atomic Model

  Is there an error in this question or solution?