1.

When a -particle of energy 3.5 MeV approaching gold nucleus, it undergoes scattering by 180^@, then distance of closest approach is :

Answer»

`12.7 xx 10^(-14)m`
33.8 fermi
`41.4 xx 10^(-14)m`
41.4 fermi

Solution :`K.E.=P.E.=(k z_(1).z_(2).e^(2))/(r_(0))`
`r_(0)=(9 xx 10^(9) xx 2 xx 79 xx (1.6 xx 10^(-19))^(2))/(5.5 xx 10^(6) xx 1.6 xx 10^(-19))m`.
`=41.4 xx 10^(-15)m`


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