0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

Anode voltage is at +3V. Incident radiation has frequency 1.4×1015Hz and work function of the photo cathode is 2.8 eV. Find the minimum and maximum KE of photo electrons in eV
  1. 3, 6
  2. 0, 3
  3. 0, 6
  4. 2.8, 5.8

A
0, 6
B
3, 6
C
0, 3
D
2.8, 5.8
Solution
Verified by Toppr

Energy of the incident radiation is:
hf=6.625×1034×1.4×10151.6×1019=5.8eV

Was this answer helpful?
1
Similar Questions
Q1
Anode voltage is at +3V. Incident radiation has frequency 1.4×1015Hz and work function of the photo cathode is 2.8 eV. Find the minimum and maximum KE of photo electrons in eV
View Solution
Q2
The voltage at anode is +3V. Incident radiation has frequency 1.4×1015Hz and work function of the photocathode is 2.8eV. The minimum and maximum kinetic energy of photo electrons in eV respectively are
View Solution
Q3
Light from a hydrogen discharge tube is incident on the cathode of a photo-cell. The work function of the cathode surface is 3.8 eV. In order to reduce the photo-electric current to zero, the voltage of anode relative to cathode must be
View Solution
Q4
On a photosensitive material, when frequency of incident radiation is increased by 30% kinetic energy of emitted photo electrons increases from 0.4 eV to 0.9 eV. The work function of the surface is :
View Solution
Q5
Photon of energy 6 eV is incident on a metal surface of work function 4 eV. Maximum KE of emitted photo-electrons will be :
View Solution