In an alpha particle scattering experiment distance of closest approach for the particle is 4.5 10-14 m. If target nucleus has atomic number 80 , then maximum velocity of -particle is 105 m/s approximately. ( 14 0 = 9 109 SI unit, mass of particle = 6.72 10-27 kg)
Answer
156
Solution
At closest approach KE → PE (head-on):
12mv2 = 14 0 (Ze)(2e)r0
v2 = 4 9 109 80 (1.6 10-19)26.72 10-27 4.5 10-14
= 2.44 1014 m2/s2
v 1.56 107 m/s = 156 105 m/s
Answer: 156