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Activation Energy

Binding energ...

Question

Binding energy per nucleon of $_{1}^{2} H$ and $_{2}^{4} H e$ are $1.1 M e V$ and $7 M e V$ respectively. Calculate the amount of energy released in the following process:
$_{1}^{2} H +_{1}^{2} H ⟶$ $_{2}^{4} H e$

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Solution

Amount of energy released $= \sum B i n d i n g e n e r g y o f p r o d u c t s - \sum B i n d i n g e n e r g y o f r e a c t a n t s$
$= [4 \times 7] - [4 \times 1.1] = 23.6 M e V$

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