A 7.9 MeV -particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) m. [ 14 0 = 9 109 N m2/C2 . and electron charge . = 1.6 10-19 C ]
- 2.88 10-14
- 5.76 10-14
- 1.44 10-13
- 1.69 10-12
Answer
(2) 5.76 10-14
Solution
The size of the nucleus is estimated from the distance of closest approach r0 (head-on collision), where the whole kinetic energy becomes electrostatic potential energy:
K = 14 0 (2e)(Ze)r0
r0 = 14 0 2Ze2K
With K = 7.9 MeV = 7.9 106 1.6 10-19 J:
r0 = 9 109 2 79 (1.6 10-19)27.9 106 1.6 10-19
r0 = 9 109 20 1.6 10-19106
r0 = 2.88 10-14 m (estimate of the radius)
Diameter = 2r0 = 5.76 10-14 m
Answer: (2)