A beam of `alpha`-particle of velocity `2.1xx10^(7)m//s` is scattered by a gold foil (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)`.

Explanation

`(1)/(2) m_(alpha) v_(alpha)^(2) = ( K(2e) (ze))/(r_(0))`
`r_(0)= ( 2K((2e)/( m_(alpha)))(79 e ))/(v_(alpha)^(2))`
`= ( 2 xx ( 9 xx 10^(8))( 4.8 xx 10^(7) ) ( 79 xx 1.6 xx 10^(-19)))/( ( 2.1 xx 10^(7))^(2))`,
`r_(0) = 2.5 xx 10^(-14) m`