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

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 : (A) 19.6 MeV (B) -2.4 MeV (C) 8.4 MeV (D) 17.3 MeV

Tardigrade Admin · Accepted Answer

BE of 2He4 = 4 × 7.06 = 28.24 MeV BE of 37Li = 7 × 5.60 = 39.20 MeV 73Li+11H → 2He4 +2He4+ Q Therefore, Q = 56.48 - 39.20 = 17.28 MeV.

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

19.6 MeV

-2.4 MeV

8.4 MeV

17.3 MeV

$BE$ of $_{2} H e^{4} = 4 \times 7.06 = 28.24 M e V$
$BE$ of $_{3}^{7} L i = 7 \times 5.60 = 39.20 M e V$
$_{3}^{7} L i +_{1}^{1} H \to_{2} H e^{4} +_{2} H e^{4} + Q$
Therefore, $Q = 56.48 - 39.20 = 17.28 M e V .$

Q. The Binding energy per nucleon of 37​Li and 24​He nuclei are 5.60MeV and 7.06MeV, respectively. In the nuclear reaction 37​Li+11​H→224​He+Q, the value of energy Q released is :