CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B

decreases with increase in E

loader
C
first increases then decreases with increase in E
loader
D
remains constant
loader

Solution

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

Co-Curriculars

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image