The energy released when frac{7}{17.13} text{ | vedclass

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(B) The nuclear reaction is: $^7_3	ext{Li} + ^1_1	ext{H} ightarrow 2(^4_2	ext{He})$.Mass of reactants $= 7.0183 	ext{ u} + 1.008 	ext{ u} = 8.0

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The nuclear reaction is: $^7_3	ext{Li} + ^1_1	ext{H} ightarrow 2(^4_2	ext{He})$.Mass of reactants $= 7.0183 	ext{ u} + 1.008 	ext{ u} = 8.0263 	ext{ u}$.Mass of products $= 2 	imes 4.004 	ext{ u} = 8.008 	ext{ u}$.Mass defect $\Delta m = 8.0263 	ext{ u} - 8.008 	ext{ u} = 0.0183 	ext{ u}$.Energy released per reaction $= 0.0183 	imes 931 	ext{ MeV} = 17.0373 	ext{ MeV} = 1.70373 	imes 10^7 	ext{ eV}$.Number of moles in $\frac{7}{17.13} 	ext{ kg}$ (i.e.,$\frac{7000}{17.13} 	ext{ g}$) of $^7_3	ext{Li}$ (molar mass $\approx 7 	ext{ g/mol}$): $n = \frac{7000 / 17.13}{7} = \frac{1000}{17.13} \approx 58.377 	ext{ mol}$.Total number of atoms $N = n 	imes N_A = 58.377 	imes 6.0 	imes 10^{23} \approx 3.5026 	imes 10^{25}$.Total energy released $= N 	imes (1.70373 	imes 10^7 	ext{ eV}) \approx 3.5026 	imes 10^{25} 	imes 1.70373 	imes 10^7 	ext{ eV} \approx 5.967 	imes 10^{32} 	ext{ eV}$.Comparing with $\alpha 	imes 10^{32} 	ext{ eV}$, we get $\alpha \approx 5.967$. The nearest integer is $6$.

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