1.

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

Answer»

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

Solution :Since KINETIC energy is converted into potentia energy
`:.` kinetic energy = potential energy
`:.5MeV=(1)/(4piepsi_(0)).(q_(1)q_(2))/(R)`
`:.5xx10^(6)xxe=(9xx10^(9)xx(92e)(2e))/(r)`
`:.r=(9xx10^(9)xx92xx2xxe)/(5xx10^(6))=5.3xx10^(-14)m`
`:.r=5.3xx10^(-12)cm`


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