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(D) Adding the two given nuclear reactions:${ }_3 Li^6 + { }_0 n^1 \rightarrow { }_2 He^4 + { }_1 H^3$${ }_1 H^2 + { }_1 H^3 \rightarrow { }_2 He^4 +

Vedclass · Accepted Answer

$22.22$

Vedclass · Answer

(D) Adding the two given nuclear reactions:${ }_3 Li^6 + { }_0 n^1 ightarrow { }_2 He^4 + { }_1 H^3$${ }_1 H^2 + { }_1 H^3 ightarrow { }_2 He^4 + { }_0 n^1$--------------------------------------------------------------${ }_3 Li^6 + { }_1 H^2 ightarrow 2({ }_2 He^4)$---------------------------------------------------------------The energy released $(Q)$ in the process is given by $Q = \Delta m 	imes 931.5 \ MeV/amu$.$\Delta m = [M(Li^6) + M({ }_1 H^2) - 2 	imes M({ }_2 He^4)]$$\Delta m = [6.01690 + 2.01471 - 2 	imes 4.00388] \ amu$$\Delta m = [8.03161 - 8.00776] \ amu = 0.02385 \ amu$$Q = 0.02385 	imes 931.5 \ MeV \approx 22.216 \ MeV$Rounding to two decimal places,$Q = 22.22 \ MeV$.

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