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

  • Option 1)


  • Option 2)

    \nu ^{2}

  • Option 3)


  • Option 4)

    1/\nu ^{4}


Answers (1)


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



For closest approach, kinetic energy is converted into potential energy

\therefore \; \; \frac{1}{2}mv^{2}=\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r_{0}}=\frac{1}{4\pi \varepsilon _{0}}\frac{(Ze)(2e)}{r_{0}}

or\; \; \; r_{0}=\frac{4Ze^{2}}{4\pi \varepsilon _{0}mv^{2}}=\frac{Ze^{2}}{\pi \varepsilon _{0}v^{2}}\left ( \frac{1}{m} \right )

or\; \; \; r_{0}\; \; is\; proportional\; to\; \left ( \frac{1}{m} \right )

Correct option is 3.

Option 1)


This is an incorrect option.

Option 2)

\nu ^{2}

This is an incorrect option.

Option 3)


This is the correct option.

Option 4)

1/\nu ^{4}

This is an incorrect option.

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