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
`10^(-12) cm`
B
`10^(-10) cm`
C
`1A`
D
`10^(-15) cm`
Text Solution
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The correct Answer is:
A
KEY CONCEPT : Distance of closest approach `r_(0) = (Ze(2e))/(4 pi epsilon _(0)E)` Energy `E = 5 xx 10^(6) xx 10^(-19) J` `:. R_(0) = (9 xx 10^(9) xx (92 xx 1.6 xx 10^(-19)) (2 xx 1.6 xx 10^(-19))/(5 xx 10^(6) xx 1.6 xx 10^(-19))` rArr r = 5.2 xx 10^(-14)m = 5.3 xx 10^(-12) cm`
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