Nuclei · Class 12 · CBSE · Science Physics

The binding energy per nucleon for {2}^{4} ext{He} is 7.07 ext{ MeV} and for {3}^{7} ext{Li} is 5.60 ext{ MeV}. Which of the following statements is…

Practice question from Nuclei (Class 12 · CBSE · Science Physics), with the correct answer and a step-by-step solution.

easy · MCQ · Q-E6824E

The binding energy per nucleon for $_{2}^{4} ext{He}$ is $7.07 ext{ MeV}$ and for $_{3}^{7} ext{Li}$ is $5.60 ext{ MeV}$ . Which of the following statements is true regarding their stability?

A. $_{3}^{7}\text{Li}$ is more stable than $_{2}^{4}\text{He}$
B. $_{2}^{4}\text{He}$ is more stable than $_{3}^{7}\text{Li}$ Correct answer
C.Both are equally stable
D.Stability cannot be compared using binding energy per nucleon

Solution

Greater the binding energy per nucleon, more stable is the nucleus. Since $7.07\text{ MeV} > 5.60\text{ MeV}$ , the helium nucleus ( $_{2}^{4}\text{He}$ ) is more stable than the lithium nucleus ( $_{3}^{7}\text{Li}$ ).

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