CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

When a certain metal was irradiated with light of frequency 4×1016 s1. The photoelectrons emitted had three times the kinetic energy as the kinetic energy of photoelectron emitted when the metal was irradiated with light of frequency2×1016 s1. If the critical frequency of the metal (νo)=x×1016 s1, then find the value of x.
(given, h=6.6×1034 Js)
.


Solution

According to Einstein photoelectric equation,
hν=hνo+KE
Where, νo is the threshold frequency
According to the question,
h×4×1016=hνo+3K.E  ...(1)
h×2×1016=hνo+K.E...(2)

Multiplying '3' in equation (2), we get
h×6×1016=3hνo+3K.E...(3)

Subtracting equation (1) from equation (3) we get
h×2×1016=2hνo
νo=1016 s1

x=1

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image