An `alpha`- particle of 10 MeV is moving forward for a head on collision. What will be the distance of closest approach from the nucleus of atomic number Z = 50 ?

Options

A. `1.44xx10^(-14)m`
B. `2.88xx10^(-14)m`
C. `0.53xx10^(-10)m`
D. `(0.53xx10^(-10))/(50)m`

Answer

A

Explanation

`1.44xx10^(-14)m`
Potential energy of a-particle at distance
d = kinetic energy of a particle at large distance
`(1)/(4piepsi_(0))((Ze)(2e))/(d)=10MeV`
`d=((Ze)(2e))/(4piepsi_(0)xx10xx10^(6)xx1.6xx10^(-19))`
`=(2xx50xx(1.6xx10^(-19))^(2))/(4xx3.14xx8.85xx10^(-12)xx10^(7)xx1.6xx10^(-19))`
`=(100xx1.6xx10^(-19))/(4xx3.14xx8.85xx10^(-5))`
`:.d=1.44xx10^(-14)m`