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An alpha nucleus of energy
1 mv2 2
bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
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1 Ze - v2
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1 m -
1 v4
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Correct Option: C
Kinetic energy of alpha nucleus is equall to electrostatic potential energy of the system of the alpha particle and the heavy nucleus. That is,
| mv2 = | ||||||
| 2 | 4πε0 | r0 |
where r0 is the distance of closest approach
| r0 = | ||||
| 4πε0 | mv2 |
| ⇒ r0 ∝ Ze ∝ qα ∝ | ∝ | = 8 | ||
| m | v2 |
Hence, correct option is (c).
