A neutron is absorbed by a ${}_{3}^{6}{Li}$ nucleus with the subsequent emission of an alpha particle.
(i) Write the corresponding nuclear reaction.
(ii) Calculate the energy released, in MeV, in this reaction.
[Given: mass ${}_{3}^{6}{Li} = 6.015126 u; mass (neutron) = 1.0086654 u$ mass (alpha particle) = 4.0026044 and mass (triton) = 3.0100000 u. Take 1 u = 931 MeV/c²].

Solution

i) Nuclear reaction is given as,

   ${}_{3}^{6}{Li} + {}_{0}^{1}n \to {}_{2}^{4}{He} + {}_{1}^{3}H + Q$

ii) Mass defect $(∆ m)$ ,

   $= [m ({}_{3}^{6}{Li}) + m ({}_{0}^{1}n) - m ({}_{2}^{4}{He}) - m ({}_{1}^{3}H)] = [6.015126 + 1.0086654 - 4.0026044 - 3.01]$

   $= [7.0237914 u - 7.0126044 u] = 0.0111870 u$

$∴$ Energy released = 0.0111870 x 931
= 10.415 MeV.

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