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  • Need clarity, kindly explain! - Atoms And Nuclei - JEE Main

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)

best_answer

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.

Posted by

perimeter

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