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An alpha - particle of 5MeV energy strik...

An alpha - particle of 5MeV energy strikes with a nucleus of uranium at stationary at an scattering angle of 180o .The nearest distance up to which alpha - particle reaches the nucleus will be of the order of

A

1 Å

B

`10^(-10)`cm

C

`10^(-12)`cm

D

`10^(-15)` cm

Text Solution

Verified by Experts

The correct Answer is:
C
`Z_(u)=92, Z_(He) = 2 `
For closest approach
`KE = 1/(4piepsilon_(0)).(q_(1)q_(2))/(r_(0))`
As `q_(1) = 92 e and q_(2)=2e`
`rArr r_(0) = (9xx10^(9)xx92 xxexx2xxe)/(5xx10^(6)xxe)`
`rArr r_(0) = 5.3 xx 10^(-14) m = 5.3 xx 10^(-12)cm `.
Here order is `10^(-12)`
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