1.

An alpha -particle of energy 5 MeV is scattered through 180^@ by a fixed uranium nuclous. The distance of closest approach is of the order:

Answer»

`10^(-10)m`
`10^(-13)m`
`10^(-14)m`
`10^(-16)m`

Solution :If `r_(0)` is the distance of closest APPROACH, K.E.=P.E.
`E=1/2 MV^(2)=((Ze)(2e))/(4PI epsi_(0) r_(0))`
`r_(0)=((Ze)2e)/(4pi epsi_(0)+E)`
`=(92 xx 1.6 xx 10^(-19) xx 2 xx 1.6 xx 10^(-19))/(5 xx 10^(6) xx 1.6 xx 10^(-19))`
`=10^(-14)m`


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