An $\alpha$ particle of energy $5MeV$ is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

Option 1)
$1\dot{A}$

Option 2)
$10^{-10}cm$

Option 3)
$10^{-12}cm$

Option 4)
$10^{-15}cm$

Answers (1)

P perimeter

As we learnt in

Binding energy per nucleon -

- wherein

This graph shows the stability of nuclei many nuclear phenomena can be explained by this graph

At distance of closest approach whole kinetic energy is converted into potential energy.

$\frac{1}{2}mv^{2}=\frac{(ze)(2e)}{4\pi \epsilon_{0}r}=5\ MeV$

$\therefore\ \; \frac{2ze^{2}}{4\pi \epsilon_{0}(5 MeV)}=5.3\times10^{-12}cm$

Correct option is 3.

Option 1)

$1\dot{A}$

This is an incorrect option.

Option 2)

$10^{-10}cm$

This is an incorrect option.

Option 3)

$10^{-12}cm$

This is the correct option.

Option 4)

$10^{-15}cm$

This is an incorrect option.

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An particle of energy is scattered through by a fixed uranium nucleus. The distance of the closest approach is of the order of Option 1) Option 2) Option 3) Option 4)

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An $\alpha$ particle of energy $5MeV$ is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

Option 1)
$1\dot{A}$

Option 2)
$10^{-10}cm$

Option 3)
$10^{-12}cm$

Option 4)
$10^{-15}cm$