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The binding energy/nucleon of deuteron $({}_{1}H^{2})$ and the helium atom $({}_{2}H e^{4})$ are 1.1MeV and 7MeV respectively. If the two deuteron atoms fuse to form a single helium atom, then the energy released is

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$26.9 M e V$

$25.8 M e V$

$23.6 M c V$

$12.9 M c V$

answer is C.

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Detailed Solution

BE of deuteron $(_{1} H^{2}) = 2 \times 1.1 M e V = 2.2 M e V$

BE of helium atom $(_{2} H e^{4}) = 4 \times 7 M e V = 28 M e V_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4} + E n e r g y r e l e a s e d$

$E n e r g y r e l e a s e d = B E o f h e l i u m - 2 \times B E o f d e u t e r o n$

$= 28 M e V - 2 \times 2.2 M e V = 28 M e V - 4.4 M e V = 23.6 M e V$

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