The energy released in the process is given by
\(Q=\left[M_{^{2}_{1} H}+M_{^{3}_{1} H}-M_{^{4}_{2} H e}-M_{^1_0n}\right] c^{2}\)
\(\;\;\;=[2.014102+3.016050-4.002603-1.008665] u c^{2}\)
\(\;\;\;=(0.018884 u)\left[931.5 \frac{ MeV }{u}\right]=17.6 \;MeV\)
\({}_1^2H + {}_1^3H \to {}_2^4H + n + 17.6\,MeV\)
(Controlled thermonuclear fission reaction)
પ્રોટોનનું દળ $m _{ P }=1.00783\, U ,$ ન્યૂટ્રોનનું દળ $m _{ n }=1.00867\, U$ અને ન્યુક્લિયસનું દળ $m _{ Sn }=119.902199$ $U.$
(લો : $1 U =931\, MeV )$
$(I)$ $_92^U{235} + _0n^1 \,X + 35^Br85 + 3 \,_0n^1$
$(II)$ $_3Li^6 + _1H^2 \,Y + _2He^4$