Eduprobe
Question:
When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photocurrent is V0/2. When the surface is illuminated by monochromatic light of frequency v2, the stopping potential is −V0. The threshold frequency for photoelectric emission is:
5v3
3v2
2v
43v
Solution:
The correct option is A
3v2
hν = W + eV0
2ehν = W + eV0
ehν2 = W − eV0
On solving we get, W = 3/2hν
hν0 = 3/2hν
ν0 = 3/2ν