An alpha nucleus of energy `(1)/(2)m nu^(2)` bombards a heavy nucleus of charge `Ze` . Then the distance of closed approach for the alpha nucleus will be prpportional to
A
`nu^(2)`
B
`(1)/(m)`
C
`(1)/(nu^(2))`
D
`(1)/(Ze)`
Text Solution
Verified by Experts
The correct Answer is:
C
Work done to step the a particle is equal to K.E `:. qV= (1)/(2) m nu^(2) rArr q xx (R(Ze))/(r ) = (1)/(2) m nu^(2)` ` rArr r = (2 (2e)k(Ze))/(m nu^(2)) = (4kZe^(2))/( m nu^(2)) ` `rArr r prop (1)/(nu^(2)) and r prop (1)/(m)`
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