CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Work function of metal A is equal to the ionization energy of hydrogen atom in the first excited state. Work function of metal B is equal to the ionization energy of He+ ion in the second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectron emitted from B.
The difference in maximum kinetic energy of photoelectrons from A and from B.


A
increases with increase in E
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

decreases with increase in E

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
first increases then decreases with increase in E
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
remains constant
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D remains constant
(EϕA)(EϕB)=ϕBϕA
=10.2 eV = constant

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Energy Levels
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon