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  • An alpha nucleus of energy  bombards a hea

An alpha nucleus of energy \frac{1}{2}\text{mv}^{2} bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

Option: 1

\frac{1}{\text{v}^{4}}


Option: 2

\frac{m}{\text{Ze}}


Option: 3

V^4


Option: 4

\frac{1}{\text{m}}


Answers (1)

best_answer

 

At distance of closest approach all kinetic energy will be converted into potential energy if distance of closest approach is b then,

\frac{1}{2}m\upsilon ^{2}=\frac{1}{4\pi\epsilon _{o} }\cdot \frac{(ze)\cdot (2e)}{b}

\Rightarrow b=\frac{(ze)\cdot (2e)}{(4\pi\epsilon_{o})\cdot (\frac{1}{2}m\upsilon ^{2})} \Rightarrow b\propto \frac {1}{m}

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