Question

If the binding energy per nucleon in ${}_7\text{3}Li$ and He nuclei are $5.60MeV$ and $7.06MeV$ respectively, then in the reaction $p{+}_3\text{7}Li{\to }_2\text{4}$ He, binding energy is

• A. 39.2 MeV
• B. 28.24 MeV
• C. 17.28 MeV
• D. 1.46 MeV

Answer 1

Answer: (C)

17.28 MeV

Since, $p{+}_{3}L{i}^7\to 2_{2}H{e}^4$

Thus, proton energy = BE of He atom $-$ BE of Li of atom.

or $E_p=(2\times 4\times 7.06)-(7\times 5.60)=17.28MeV$ where 4 is the no of nucleon of He and 7 is the no of nucleon of Li.