The Q-value of a nuclear reaction A + b → C + d is defined by
Q = [m_A + m_b - m_c - m_d] c²where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
(i) ${}_{1}^{1}H + {}_{1}^{3}H \to {}_{1}^{2}H + {}_{1}^{2}H$
(ii) ${}_{6}^{12}C + {}_{6}^{12}C \to {}_{10}^{20}{Ne} + {}_{2}^{4}{He}$
Atomic masses are given to be
$m ({}_{1}^{2}H) = 2.014102 u; m ({}_{1}H^{3}) = 3.0160409 u m ({}_{6}^{12}C) = 12.000000 u; m ({}_{10}{Ne}^{20}) = 19.992439 u .$

Solution

(i) Considering the first reaction,

   ${}_{1}^{1}H + {}_{1}^{3}H \to {}_{1}^{2}H + {}_{1}^{2}H$

Q-value is given by,
$Q = ∆ m \times 931.5 MeV = [m ({}_{1}^{1}H) + m ({}_{1}^{3}H) - 2 m ({}_{1}^{2}H)] \times 931 MeV = [1.007825 + 3.016049 - 2 \times 2.014102] \times 931 Mev = - 4.03 MeV$
Since, Q-value is negative, this reaction is endothermic.

(ii) The second reaction is,

   ${}_{6}^{12}C + {}_{6}^{12}C \to {}_{10}^{20}{Ne} + {}_{2}^{4}{He}$

Q-value is given by,
   $Q = ∆ m \times 931 MeV = [2 m ({}_{6}^{12}C) - m ({}_{10}^{20}{Ne}) - m ({}_{2}^{4}{He})] \times 931 MeV = [24.000000 - 19.992439 - 4.002603] \times 931 MeV = + 4.61 MeV$

Since, the Q-value is positive , the reaction is exothermic.

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