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  4. A neutron is absorbed by a 36 L i nucleus with the subsequent emission of an alpha particle. 36 mathrmLi+ 01 n arrow 24 mathrmHe+ 13 mathrmH+Q Calculate the energy released, in MeV , in this reaction. [Given: mass 36 textitLi=6.015126 u; mass (neutron) =1.0086654 u; Mass (alpha particle) =4.0026044 u and Mass (tritium) =3.0100000 u. [ Take 1 u=931 MeV c- 2 ]

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Q. A neutron is absorbed by a $_{3}^{6} L i$ nucleus with the subsequent emission of an alpha particle.

${ }_3^6 \mathrm{Li}+{ }_0^1 n \rightarrow{ }_2^4 \mathrm{He}+{ }_1^3 \mathrm{H}+Q$

Calculate the energy released, in $MeV$ , in this reaction.

[Given: mass $_{3}^{6}\textit{Li}=6.015126 \, u;$ mass (neutron) $=1.0086654 \, u;$

Mass (alpha particle) $=4.0026044 \, u$ and

Mass (tritium) $=3.0100000 \, u.$

[ Take $1 \, u=931 \, MeV \, c^{- 2}$ ]

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Solution:

${ }_3^6 \mathrm{Li}+{ }_2^4 n \rightarrow{ }_2^4 \mathrm{He}+{ }_1^3 \mathrm{H}+Q$
Total initial mass $= 6.015126 + 1.0086654$
$= 7.0237914 \, \text{amu}$
Total final mass $= 4.0026044 + 3.01$
$= 7.0126044 \, \text{amu}$
Mass defect, $\Delta m = 7.0237914 - 7.0126044$
$= 0.0111870 \, \text{amu}$
Energy released, $Q = 0.0111870 \times 931$
$=10.415 \, MeV.$