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Activation Energy

The binding e...

Question

The binding energy of $_{2}^{4} H e$ is $28.57$ $M e V$ . What shall be the binding energy per nucleon of this element?

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Solution

The nucleus of $_{2}^{4} H e$ consists of $4$ nucleons.
So, Binding energy per nucleon $= \frac{T o t a l b i n d i n g e n e r g y}{N u m b e r o f n u c l e o n s}$
$= \frac{28.57}{4} = 7.14 M e V$

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