Class 12 NEET Physics Atoms and Nuclei Questions Part 2

At the distance of closest approach d. Kinetic energy = Potential energy

1

2

mv2=

1

4πε0

(2e)(Ze)

d

where, Ze = charge of target nucleus
2e = charge of alpha nucleus

1

2

mv2=kinetic energy of alpha nucleus of mass m moving with velocity v
or d=

2Ze2

4πε0( 1 2 mv2)

∴d∝

1

m