A neutron is absorbed by a ${^6_3Li}$ nucleus with subsequent emission of an alpha particle. The energy released in the reaction is
Use mass of ${^6_3Li} = 6.015126 u$
Mass of ${^4_2He} = 4.002604 u$
Mass of ${^1_0n} = 1.008665 u$
Mass of ${^3_1H} = 3.016049 u$ and $1 u = 931 MeV$
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4.78 Mev
The reaction is:
$^{6}_{3}\text{Li} + ^{1}_{0}\text{n} \rightarrow ^{4}_{2}\text{He} + ^{3}_{1}\text{H}$
Masses given:
$m(^{6}\text{Li}) = 6.015126 \, \text{u}$
$m(^{1}\text{n}) = 1.008665 \, \text{u}$
$m(^{4}\text{He}) = 4.002604 \, \text{u}$
$m(^{3}\text{H}) = 3.016049 \, \text{u}$
Mass of reactants: $6.015126 + 1.008665 = 7.023791 \, \text{u}$
Mass of products: $4.002604 + 3.016049 = 7.018653 \, \text{u}$
Mass defect: $\Delta m = 7.023791 - 7.018653 = 0.005138 \, \text{u}$
Energy released: $E = \Delta m \times 931 \, \text{MeV} = 0.005138 \times 931 \approx 4.78 \, \text{MeV}$