The $Q$ value of a nuclear reaction $A+b \rightarrow C+d$ is defined by $Q=\left[m_{A}+m_{b}-m_{C}-m_{d}\right] c^{2}$ where 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.
$(i) \;_{1}^{1} H+_{1}^{3} H \rightarrow_{1}^{2} H+_{1}^{2} H$
$(ii)\;_{6}^{12} C+_{6}^{12} C \rightarrow_{10}^{20} N e+_{2}^{4} H e$
Atomic masses are given to be:
$m(_{1}^{1}H) = 1.007825 \; u$
$m(_{1}^{2}H) = 2.014102 \; u$
$m(_{1}^{3}H) = 3.016049 \; u$
$m(_{6}^{12}C) = 12.000000 \; u$
$m(_{10}^{20}Ne) = 19.992439 \; u$
$m(_{2}^{4}He) = 4.002603 \; u$

(N/A) $(i)$ The given nuclear reaction is: $_{1}^{1} H+_{1}^{3} H \rightarrow_{1}^{2} H+_{1}^{2} H$
Using the formula $Q = [m(_{1}^{1}H) + m(_{1}^{3}H) - 2m(_{1}^{2}H)]c^{2}$:
$Q = [1.007825 + 3.016049 - 2(2.014102)] \; u \cdot c^{2}$
$Q = [4.023874 - 4.028204] \; u \cdot c^{2} = -0.00433 \; u \cdot c^{2}$
Since $1 \; u \cdot c^{2} = 931.5 \; MeV$,$Q = -0.00433 \times 931.5 \approx -4.033 \; MeV$.
As $Q < 0$,the reaction is endothermic.
$(ii)$ The given nuclear reaction is: $_{6}^{12} C+_{6}^{12} C \rightarrow_{10}^{20} N e+_{2}^{4} H e$
Using the formula $Q = [2m(_{6}^{12}C) - m(_{10}^{20}Ne) - m(_{2}^{4}He)]c^{2}$:
$Q = [2(12.000000) - 19.992439 - 4.002603] \; u \cdot c^{2}$
$Q = [24.000000 - 23.995042] \; u \cdot c^{2} = 0.004958 \; u \cdot c^{2}$
$Q = 0.004958 \times 931.5 \approx 4.618 \; MeV$.
As $Q > 0$,the reaction is exothermic.