The binding energy per nucleon of 37 Li and 24 He nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction 37 Li + 11 H arrow 2 24 He +Q, the value of energy Q (in MeV ) released is

Question

The binding energy per nucleon of 37 Li and 24 He nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction 37 Li + 11 H arrow 2 24 He +Q, the value of energy Q (in MeV ) released is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

BE of 2 He 4=4 × 7.06=28.24 MeV BE of 37 Li =7 × 5.60=39.20 MeV underset39.20 37 Li + 11 H longrightarrow underset28.24×2 2 He 4+ 2 He 4+Q 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 $Q$ (in $M e V$ ) released is

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

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 (in MeV ) released is

BE of 2​He4=4×7.06=28.24MeV BE of 37​Li=7×5.60=39.20MeV 39.2037​Li​+11​H⟶28.24×22​He4​+2​He4+Q Q=56.48−39.20=17.28MeV

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 $Q$ (in $M e V$ ) released is

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