Question

The binding energy per nucleon of $Li$ and $He$ nuclei are $5.60\, MeV$ and $7.06\, MeV$ respectively. In the nuclear reaction ${}^7_3Li+{}_1^{1}H \rightarrow \, \, {}_2^4He +{}_2^4He+Q$ the value of energy Q released is

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

Answer 1

Answer: (D)

Binding energy of ${}_{3}^{7}Li$ nucleus
$=7 \times 5.60\, MeV=39.2\, MeV$
Binding energy of ${}_{2}^{4} He$ nucles
$=4 \times 7.06\, MeV=28.24\, MeV$
The reaction is
$_{3}^{7}Li + {}_{1}^{1}H \rightarrow \, 2(_2^4 He)+Q$
$\therefore Q=2$ (BE of ${}_{2}^{4}He$)-(BE of ${}_{3}^{7}Li$)
$=2 \times 28.24\, MeV-39.2\, MeV$
$=56.48\, MeV-39.2\, MeV$
$=17.28\, MeV$