Would the energy be released or needed for the following D-T reaction
${}_{1}^{2}H + {}_{1}^{3}H \to {}_{2}^{4}{He} + {}_{0}^{1}n$ to occur?
Given:
$m ({}_{1}^{2}H) = 2.014102 u, m ({}_{1}^{3}H) = 3.016049 u$
$m ({}_{2}^{4}{He}) = 4.002603 u, m ({}_{0}^{1}n) = 1.008665 u$
and $1 u = 931 MeV / c^{2}$
Calculate this energy in MeV.

Solution

We have,

Mass $({}_{1}^{2}H)$ + mass $({}_{1}^{3}H)$ = 2.014102 + 3.016049
= 5.030151 u
and
$m ({}_{2}^{4}{He}) + m ({}_{0}^{1}n) = 4.002603 + 1.008665 = 5.011268 u$

As seen from above, since, mass of the reactants is larger than mass of the products therefore, energy is released in this reaction.

Now,

$Mass defect (∆ m) = 5.030151 - 5.011268 = 0.018863 u$

So, energy required = $(∆ m) \times 931 MeV$

$= 0.018863 \times 931 MeV = 17.561453 MeV = 17.56 MeV .$

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