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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 8 \times 10^{6}
Option: 2 1.33 \times 10^{6}
Option: 3 4.5 \times 10^{5}
Option: 4 6.82 \times 10^{5}

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\\ \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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Deependra Verma

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