Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Photons of energies $1\; eV$ and $2\; eV$ are successively incident on a metallic surface of work function $0.5\; eV$ . The ratio of kinetic energy of most energetic photoelectrons in the two cases will be.
Option: 1 $1:2$
Option: 2 $1:1$
Option: 3 $1:3$
Option: 4 $1:4$

Answers (1)

The ratio of kinetic energies are

$KE=h\nu-work\ function$

$=\frac{1-0.5}{2-0.5}=\frac{0.5}{1.5}=\frac{1}{3}$

Posted by

Safeer PP

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE extends single girl child scholarship registration deadline till November 20

Latest Question