$_{1} H^{1} +_{3} L i^{7} \to$ ______ $+_{2} H e^{4}$

_{2} H e^{4}

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_{1} H^{1}

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_{4} B e^{9}

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_{4} B e^{8}

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Solution

The correct option is A $_{2} H e^{4}$
$_{1} H^{1} +_{3} {L i}^{7} ⟶_{2} {H e}^{4} +_{2} {H e}^{4}$
By law of conservation in nuclear reactions
Conservation of mass number and charge number
Before the reaction
Mass number (A) $= 1 + 7$
$= 8$
Charge number (z) $= 1 + 3 = 4$
After the reaction
Mass number (A) $= 4 + 4$
$= 8$
Charge number (z) $= 2 + 2 = 4$

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