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An alpha particle of energy 5 MeV is sca...

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

A

`1 overset(@)(A)`

B

`10^(-10)`cm

C

`10^(-12)cm`

D

`10^(-15)cm`

Text Solution

Verified by Experts

The correct Answer is:
C
K.E. of `alpha-`particle = P.E. of electrostatic system
`rArr" "5MeV=1/(4piepsilon_(0))(q_(1)q_(2))/r=((9 times 10^(9)) times (Z_(1)e)(Z_(2)e))/r`
`rArr 5 times 10^(6) times e=(9 times 10^(9)) times (Z_(1) times Z_(2) times e^(2))/r`
`rArr r=((9 times 10^(9)) times 92 times 2 times 1.6 times 10^(-19))/(5 times 10^(6))`
`rArr r=5.3 times 10^(-14)m=5.3 times 10^(-12)cm`
`rArr r approx 10^(-12)cm`
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