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In the nuclear reaction $_{1} H^{2} +_{1} H^{2} \to_{2} {He}^{3} +_{0} n^{1}$ If the mass of the deuterium atom=2.014741 amu, mass of $_{2} {He}^{3}$ atom =3.016977 amu and mass of neutron=1.008987 amu, then the Q value of the reaction is nearly.

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2.45 MeV

3.27 MeV

0.00352MeV

0.82 MeV

answer is B.

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Detailed Solution

$Q = (\sum m_{r} - \sum m_{p}) C^{2}$

$\sum m_{r} = 2 \times 2 .014741 amu$

$\sum m_{p} = (3 .016977 + 1 .008987) amu$

$Δ m = 4 .029482 - 4 .025964 = 0 .003518 amu$

$Q = 0 .003518 \times 931 MeV = 3 .27 MeV$

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