The binding energy per nucleon of $^7_3Li$ and $^4_2He$ nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction $^7_3Li + ^1_1H \rightarrow ^4_2He + ^4_2He + Q$, the value of energy Q released is :

Question

Dr. Abubacker · Accepted Answer

[Solution] Q = 8 × 7.06 − 7 × 5.60 = 56.48 − 39.20 ≈ 17.3 MeV. Answer (4).

Q. The binding energy per nucleon of $^7_3Li$ and $^4_2He$ nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction $^7_3Li + ^1_1H \rightarrow ^4_2He + ^4_2He + Q$, the value of energy Q released is :

Explanation

Reviewed by Dr. Abubacker

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