An alpha particle of energy 5 MeV is scattered through `180^(@)` by a fixed uranium nucleus. The distance of closest approach is of the order of

Explanation

Accroding to law of convation of energy kinetic energy of `alpha` =potential energy of `alpha` particle energy of a `alpha` particle at distance of closet approach
`i.e., " " (1)/(2)mv^(2)=(1)/(4piepsi_(0))(q_(1)q_(2))/(r)`
Given atomic of uranium , Z=92 ,
Charge of electron `e=1.6xx10^(-19)C,`
Charge of `alpha` particle , `q_(1)=2e`
Charge of uranium nueleus,
`q_(2)=92e`
` :. 5 MeV=(9xx10^(9)xx(2e)xx(92e)
)/(r) " " ( :.(1)/(2)mv^(2)=5MeV)`
`rArrr=(9xx10^(9)xx2xx92xx(1.6xx10^(-19))^(2))/(5xx10^(6)xx1.6xx10^(-19))" " ( :. MeV=1.6xx10^(-19)J)`
`:." " r=5.3xx10^(-14)m`
`=5.3xx10^(-12) cm`