Determine the distance of closest approach when an alpha particle of kinetic energy 4.5 MeV strikes a nucleus of Z = 80, stops and reverses its direction.

Explanation

Let r be the center to centre distance between the alpha particle and the nucleus ( Z = 80 ) . When the alpha particle is at the stopping point, then
`K =( 1)/( 4pi epsilon_(0)) ((Ze ) ( 2e))/( r )`
( or ) `r = ( 1)/( 4pi epsilon_(0)). ( 2Ze^(2))/( K )`
`= ( 9 xx 10^(9) xx 2 xx 80 e^(2))/( 4.5 MeV)`
`= ( 9 xx 10^(9) xx 2 xx 80 xx ( 1.6 xx 10^(-19))^(2))/( 4.5 xx 10^(6) xx 1.6 xx 10^(-19) J)`
`= ( 9 xx 160 xx 16)/( 4.5 ) xx 10^(-16) = 512 xx 10^(-16) m`
`= 5.12 xx 10^(-14) m`.