1.

An alpha-particle of energy 1/2 mv^(2) bombards a heavy nuclear target of charge Ze. Then distance of closest approach for a nucleus is proportional to :

Answer»

`1/(Ze)`
`v^(2)`
`1/m`
`1/v^(4)`

Solution :`1/2 MV^(2)=(1)/(4pi epsi_(0)) (2.e Ze)/(r_(0))`
`r_(0)=(1)/(4pi epsi_(0))(4Ze^(2))/(mv^(2))`
`r_(0) ALPHA 1/m`


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