# You are given that mass of  $^{7}_{3}Li = 7.0160 \mu$  mass of  $^{4}_{2}He= 4.0020 \mu$  mass of  $^{1}_{1}H= 1.0079 \mu$ When 20g of  $^{7}_{3}Li$ is converted into $^{4}_{2}He$  by proton capture, the energy liberated (in kWh) , is:- (mass of nucleus $= 1 \frac{GeV}{C^{2}}$) Option: 1 Option: 2 Option: 3 Option: 4

$\\ \text{The nuclear reaction is}\\ { }_{3} L i^{7}+\ _{1} H^{1} \rightarrow 2\ { }_2 \mathrm{He}^{4}+Q \\ Mass defect,\ \Delta m=m_{Li}+m_{p}+2m_{He}\\ \Rightarrow \Delta m=7.0160+1.0079-2 \times 4.0026 =0.0187 a . m . u \\ Q=0.0187 \times 931 \mathrm{MeV}=17.40\ \mathrm{MeV} \\ Number of nuclei in 20 \mathrm{g} of \ 3 \mathrm{Li}^{7} =\frac{6.023 \times 10^{23}}{7} \times 20 =1.72 \times 10^{24} \\ Total energy released when 20 \mathrm{g} of { }_{3} \mathrm{Li}^{7} is converted into Helium,\\ Q =17.40 \times 1.72 \times 10^{24} \mathrm{MeV} =17.40 \times 1.72 \times 1.6 \times 10^{-13} \times 10^{24} J \\ As 1 k W h=3.6 \times 10^{6} J \\ \therefore E=\frac{17.40 \times 1.72 \times 1.6 \times 10^{11}}{3.6 \times 10^{6}} k W h\\ E=1.33 \times 10^{6} \mathrm{kWh}$

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