When two monochromatic lights of frequency, and 2 are incident on a photoelectric metal, their stopping potential becomes Vs2 and Vs respectively. The threshold frequency for this metal is
- 2
- 3
- 23
- 32
Answer
(4) 32
Solution
Einstein's photoelectric equation: eV0 = h - h 0
For frequency : e Vs2 = h - h 0 ...(i)
For frequency 2: eVs = h 2 - h 0 ...(ii)
From (i) 2 = (ii):
2h - 2h 0 = h 2 - h 0
h 0 = 2h - h 2 = 32h
0 = 32
Answer: (4)