If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is m. (Atomic number of gold = 79 and 14 0 = 9 109 in SI units)
- 2.95 10-16
- 3.85 10-14
- 2.95 10-14
- 3.85 10-16
Answer
(3) 2.95 10-14
Solution
At the distance of closest approach r0, the whole kinetic energy changes into electrostatic potential energy:
K = 14 0 (2e)(Ze)r0
r0 = 14 0 2Ze2K, with K = 7.7 106 1.6 10-19 J
r0 = 9 109 2 79 (1.6 10-19)27.7 106 1.6 10-19
r0 = 9 109 158 1.6 10-197.7 106
r0 2.95 10-14 m
Answer: (3)