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Structure of an Atom

In the nuclea...

Question

In the nuclear reaction $_{1} H^{2} +_{1} H^{2} + \to_{2} {H e}^{3} +_{0} n^{1},$ Mass of deuterium atom $= 2.014741 a m u,$ Mass of $_{2} H e^{3} atom = 3.016977 a m u,$ and Mass of neutron $= 1.008987 a m u .$ The Q-value of the reaction will be

0.00352 M e V

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3.27 M e V

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0.82 M e V

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2.45 M e V

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Solution

The correct option is B $3.27 M e V$
$\begin{matrix} Q = (Δ m) c^{2} Δ m \to M ass defect c^{2} \to 931.5 M e v Δ m = m_{r} - m_{p} \end{matrix}$

$\begin{matrix} Q = (2 \times 2.014741 - 3.016947 - 1.008987) c^{2} Q = (4.025964) 931.5 M e V Q = 3.27 M e v \end{matrix}$

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Similar questions

_{1} H^{2} +_{1} H^{2} \to_{2} H e^{3} +_{0} n^{1}

= 2.014741 a m u

_{2} H e^{3}

= 3.016977 a m u

= 1.008987 a m u

_{1} H^{2} +_{1} H^{3} \to_{2} {He}^{4} +_{0} n^{1}

_{1} H^{2} = 2.014,_{1} H^{3} = 3.016

_{2} {He}^{4} = 4.003,_{0} n^{1} = 1.009 amu

_{1} H^{2} +_{1} H^{2} ⟶_{2} {H e}^{4} +_{0} n^{1}

2.05

3.017

1.009

1 K g

1 a m u = 931.5 M e V / c^{2}

H^{2} + H^{2} \to H e^{3} + X,

X

_{1} H^{2} +_{1} H^{2} ⟶_{2} H e^{4}

200 M W

25

_{1} H^{2}

_{2} {H e}^{4}

2.0141 a m u

4.0026

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