1.

An particle of energy 5 MeV is scattered through 180° by gold nucleus. The distance of closest approach is of the order of

Answer»

`10^(-16)cm`
`10^(-14)cm`
`10^(-12)cm`
`10^(-10)cm`

Solution :`D= (1)/(4pi epsi_(0)) ((Ze) (2E))/(K) RARR D= (9 xx 10^(9) xx 7.9 xx 1.6 xx 10^(-19) xx 2 xx 1.6 xx 10^(-19))/(5 xx 10^(6) xx 1.6 xx 10^(-19))`
`D= 455 xx 10^(-16) = 455 xx 10^(-14) cm= 4.55 xx 10^(-12) cm ~~ 10^(-12)cm`


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