wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The graph shown below depicts plot of photocurrent versus anode potential for a cathode with 4 eV work function. The energy of the incident photon is
1068243_c78a16c0021249e4975ccd3e8667697f.png

A
6 eV
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
4 eV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2 eV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
8 eV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 6 eV

It is given that

Work function

ϕ=4eV

V=2V

From Einstein’s relation

hν=ϕ+Ek

Where hν is the energy of photons. And Ek is the kinetic energy

hν=(4+2)eV

hν=6eV


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Gauss' Law Application - Electric Field Due to a Sphere and Thin Shell
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon