1.

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

Answer»

`10^(-15)cm`
`10^(-12)cm`
`10^(-10)cm`
1Å

Solution :`r_(0)=1/(4pi epsi_(0)) .(q_(1).q_(2))/(E)=(1)/(4pi epsi_(0)) .(2E .92e)/(5 XX 1.6 xx 10^(-13))`
`=(9 xx 10^(9) xx 2 xx 92 xx 1.6 xx 10^(-19) xx 1.6 xx 10^(-19))/(5 xx 1.6 xx 10^(-13))`
`approx 10^(-12) m`


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