A beam of `alpha`-particle of velocity `2.1xx10^(7)m//s` is scattered by a gold ofil (Z=79). Find the distance of closest approach of `alpha`- particle to the gold nucleus. for `alpha`-particle, `2e//m=4.8xx10^(7)kg^(-1)`.

Answer

`2.5xx10^(-14)m`

Explanation

Here, `v=2.1xx10^(7)m//s, Z=79`,
`r_(0)=? (2e)/m=4.8xx10^(7)Ckg^(-1)`.
As `r_(0)=1/(4pi in_(0)) ((Ze)(2e))/(1/2 mv^(2)) = 1/(4pi in_(0)) ((2Ze)((2e)/m))/(v^(2))`
`=(9xx10^(9)xx2xx79xx1.6xx10^(-19)(4.8xx10^(7)))/((2.1xx10^(7))^(2))`
`r_(0)=2.5xx10^(-14)m`