When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is m. (h = 6.63 10-34 J s, e = 1.6 10-19 C, c = 3 108 m/s)
- 2.9 10-8
- 2.5 10-7
- 3.1 10-7
- 2.2 10-8
Answer
(2) 2.5 10-7
Solution
Einstein's equation: eV0 = hc - 0
First light: hc - 0 = 3.2 eV ...(i)
Second light: hc2 - 0 = 0.7 eV ...(ii)
(i) − (ii): hc2 = 2.5 eV hc = 5 eV
= hc5 eV = 6.63 10-34 3 1085 1.6 10-19
= 19.89 10-268 10-19 = 2.49 10-7 m
2.5 10-7 m
Answer: (2)