1. Tardigrade
  2. Question
  3. Physics
  4. The Binding energy per nucleon of 73Li and 42He nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction 73Li+11H → 242He + Q, the value of energy Q released is :

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The Binding energy per nucleon of $^{7}_{3}Li$ and $^{4}_{2}He$ nuclei are $5.60\, MeV$ and $7.06\, MeV$, respectively.
In the nuclear reaction $^{7}_{3}Li+^{1}_{1}H \to 2^{4}_{2}He + Q,$
the value of energy $Q$ released is :

VITEEEVITEEE 2018

Solution:

$BE $ of $_{2}He^{4} = 4 \times 7.06 = 28.24\, MeV$
$BE$ of $_{3}^{7}Li = 7 \times 5.60 = 39.20 \,MeV$
$^{7}_{3}Li+^{1}_{1}H \to _{2}He^{4} +_{2}He^{4}+ Q$
Therefore, $Q = 56.48 - 39.20 = 17.28\,MeV.$