Given that the mass of { }_{3}^{7} Li = 7.016 | vedclass

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(B) The nuclear reaction is: ${ }_{3}^{7} Li + { }_{1}^{1} H \rightarrow 2 { }_{2}^{4} He$.The mass defect $\Delta m$ for one reaction is: $\Delta m =

Vedclass · Accepted Answer

$1.33 	imes 10^{6}$

Vedclass · Answer

(B) The nuclear reaction is: ${ }_{3}^{7} Li + { }_{1}^{1} H ightarrow 2 { }_{2}^{4} He$.The mass defect $\Delta m$ for one reaction is: $\Delta m = (m_{Li} + m_{H}) - 2(m_{He}) = (7.0160 + 1.0079) - 2(4.0026) = 8.0239 - 8.0052 = 0.0187 \, u$.The energy released per reaction is $Q = \Delta m 	imes 931.5 \, MeV = 0.0187 	imes 931.5 \approx 17.42 \, MeV$.Number of $Li$ atoms in $20 \, g$: $N = \frac{20}{7} 	imes 6.022 	imes 10^{23} \approx 1.72 	imes 10^{24}$ atoms.Total energy $E = N 	imes Q = (1.72 	imes 10^{24}) 	imes (17.42 	imes 10^6 	imes 1.6 	imes 10^{-19} \, J) \approx 4.79 	imes 10^{12} \, J$.Converting to $kWh$: $E_{kWh} = \frac{4.79 	imes 10^{12}}{3.6 	imes 10^6} \approx 1.33 	imes 10^6 \, kWh$.

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