In Rutherford experiment, a 5.3 MeV `alpha`-particle moves towards the gold nucleus (Z=79). How close does the `alpha`-particle to get the centre of the nucleus, before it comes momentarily to rest and reverses its motion? (`epsi_(0)=8.8xx10^(-12)F//m`)

Options

A. `3.4xx10^(-15)m`
B. `8.6xx10^(-14)m`
C. `4.3xx10^(-14)m`
D. `1.6xx10^(-14)m`

Answer

C

Explanation

The kinetic energy of `alpha`-particles is completely converted into potential energy is given by
`(1)/(2)m u^(2)=(1)/(4piepsi_(0))((2e)(Ze))/(r_(0))`
`therefore`Distance closest approch,
`r_(0)=(1)/(4piepsi_(0))(2Ze^(2))/((1)/(2)m u^(2))=(9xx10^(9)xx2xx79xx(1.6xx10^(-19))^(2))/(5.3xx1.6xx10^(-19)xx10^(6))`
`=(9xx2xx79xx1.6xx10^(9)xx10^(-38))/(5.3xx1.6xx10^(-19)xx10^(6))`
`=(9xx2xx79xx1.6xx1.6xx10^(9)xx10^(-38))/(5.3xx1.6xx10^(-19)xx10^(6))=4.3xx10^(-14)m`