An `alpha`-particle of energy 5 MeV is moving forward for a head on collision. The distance of closest approach from the nucleus of atomic numbe Z=50 is ... `xx10^(-14)m (k=9xx10^(9)SI, c=1.6xx10^(-19)C)`,

Options

A. `0.72`
B. `2.88`
C. `1.44`
D. `5.76`

Answer

B

Explanation

Kinetic energy of a-particle = Potential energy of a-particle at very large distance,
`5xx10^(6)xx1.6xx10^(-19)=(k(2Ze^(2)))/(d)`
`:.d=(k(2Ze^(2)))/(8xx10^(-13))`
`:.d=(230.4xx10^(-28))/(8xx10^(-13))`
`:.d=28.8xx10^(-5)m:d=2.88xx10^(-14)m`