Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be
- 1 : 4
- 1 : 2
- 1 : 1
- 1 : 5
Answer
(2) 1 : 2
Solution
Kmax = h - 0
K1 = 1 - 0.5 = 0.5 eV, K2 = 2.5 - 0.5 = 2.0 eV
K = 12mv2 v K
v1v2 = 0.52.0 = 12
v1 : v2 = 1 : 2
Answer: (2)