Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A monochromatic radiation of wavelength  is incident on a stationary atom as a result of which the wave

A monochromatic radiation of wavelength \lambda_1 is incident on a stationary atom as a result of which the wavelength of the photon after the collision becomes \lambda_2 and the recoiled atom has De Broglie's wavelength \lambda_3. Then,

Option: 1

\lambda_3=\sqrt{\lambda_1 \lambda_2}


Option: 2

\lambda_2=\frac{\lambda_2 \lambda_3}{\lambda_2+\lambda_3}


Option: 3

\lambda_1=\sqrt{\lambda_1^2+\lambda_2^2}


Option: 4

\lambda_3=\sqrt{\lambda_1^2+\lambda_2^2}


Answers (1)

best_answer

Conservation of momentum yields

\frac{\mathrm{h}}{\lambda_1}+0=\frac{\mathrm{h}}{\lambda_2}+\mathrm{mv}                    \Rightarrow \frac{\mathrm{h}}{\lambda_1}-\frac{\mathrm{h}}{\lambda_2}=\mathrm{mv}

\Rightarrow \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{\mathrm{mv}}{\mathrm{h}}                              ...(1)

\text { Since, } \frac{\mathrm{h}}{\mathrm{mv}}=\lambda_3                           \Rightarrow \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}

\Rightarrow \frac{1}{\lambda_1}=\frac{1}{\lambda_2}+\frac{1}{\lambda_3}

\Rightarrow \lambda_1=\frac{\lambda_2 \lambda_3}{\lambda_2+\lambda_3}

Posted by

rishi.raj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Mains 2025 Paper 2 Answer Key LIVE: BArch final answer key PDF expected soon at jeemain.nta.nic.in
anam-khan

JEE Main 2025 response sheet errors: Delhi HC allows student to apply for JEE Adv, seeks NTA explanation
anam-khan

Four Kashmiri girls defy the odds, score 99+ percentile in JEE Main 2025

Latest Question