Question

The binding energy per nucleon of ${}_3^{7}Li$ and ${}_2^{4}He$ nuclei are 5.60 MeV and 7.06 Me V, respectively. In the nuclear reaction ${}_3^{7}Li{+}_1^{1}He{⟶}_2^{4}H{+}_2^{4}He+Q$ the value of energy Q released is

• A. -2.4 MeV
• B. 8.4 MeV
• C. 17.3 MeV
• D. 19.6 MeV

Answer 1

The correct option is C 17.3 MeV

The binding energy for ${}_1{H}^1$ is around zero and also not give in the question so we can ignore it,

$Q=2(4\times 7.06)-7\times \left(5.60\right)$

$=(56.48-39.2)MeV$

$=17.28MeV$

$=17.3MeV$