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Nuclear Fusion

4 1H1 → 2...

Question

$4$ $_{1} H^{1} \to$ $_{2} H e^{4} + 2$ $_{+ 1} e^{0} + e n e r g y$
Energy released in this process is [It is given 4 $_{1} H^{1} =$ 4.031300 amu, $_{2} H e^{4} =$ 4.0026603 amu, 2 $_{1} e^{0} =$ 0.01098 amu]

A

14 M e v

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B

16.45 M e v

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C

37.2 M e v

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D

32.7 M e v

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Solution

The correct option is B $16.45 M e v$
Mass at reactant side $= 4.0313 a m u$
Mass at product side $= 4.0026603 + 0.01098 a m u$
Mass defect $= 0.0176597 a m u$
So, energy released $= 931.5 \times 0.0176597 M e V$
$= 16.45 M e V$

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