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  • An alpha nucleus of energy  bombards a heavy nuclear target of charge&nbs

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

Option: 1

\frac{1}{Z_{\ell}}


Option: 2

v^2


Option: 3

\frac{1}{m}


Option: 4

\frac{1}{v^2}


Answers (1)

best_answer

The kinetic energy of the alpha nucleus is equal to the electrostatic potential energy of the system of
the alpha particles and the heavy nucleus. That is,

\frac{1}{2} \mathrm{~m} v^2=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_\alpha \mathrm{Z}_{\ell}}{\mathrm{m} v^2}

Where r_{0} is the distance of the closest approach

\begin{aligned} & \mathrm{r}_0=\frac{2}{4 \pi \varepsilon_0} \frac{\mathrm{q}_\alpha \mathrm{Z}_{\ell}}{\mathrm{m} v^2} \\ & \Rightarrow \mathrm{r}_0 \propto \mathrm{Z}_{\ell} \propto \frac{1}{\mathrm{~m}} \propto \frac{1}{v^2} \end{aligned}

Posted by

Kuldeep Maurya

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