A 5 watt source emits monochromatic light of wavelength 5000 A. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m, the number of photoelectrons liberated will be reduced by a factor of
- 8
- 16
- 2
- 4
Answer
(4) 4
Solution
As per NCERT, the photoelectric current (number of photoelectrons emitted per second) is directly proportional to the intensity of incident light.
For a point source, I 1r2
N1N2 = I1I2 = ( r2r1 )2 = ( 1.00.5 )2 = 4
The number of photoelectrons is reduced by a factor of 4.
Answer: (4)