The minimum frequency of photon required to break a particle of mass 15.348 amu into 4 particles is kHz. [mass of He nucleus = 4.002 amu, 1 amu = 1.66 10-27 kg, h = 6.6 10-34 J s and c = 3 108 m/s]
- 14.94 1020
- 9 1019
- 9 1020
- 14.94 1019
Answer
(4) 14.94 1019
Solution
The mass of the products is 4 4.002 = 16.008 amu, so the increase in mass is m = 16.008 - 15.348 = 0.660 amu.
This mass must be supplied by the photon energy, so h min = m c2, where
m c2 = 0.66 1.66 10-27 (3 108)2 = 9.86 10-11 J
min = 9.86 10-116.6 10-34 = 1.494 1023 Hz
min = 1.494 1020 kHz = 14.94 1019 kHz
Answer: (4)