The binding energy per nucleon for the parent nucleus is $\mathrm{E}_{1}$ and that for the daughter nuclei is $\mathrm{E}_{2}$. Then

(B) $\mathrm{E}_{1}>\mathrm{E}_{2}$

JEE Main · 2010 · 11 Apr 2010 · Offline · Modern Physics

Passage

A nucleus of mass

M + Δ m

is at rest and decays into two daughter nuclei of equal mass

\frac{M}{2}

each. Speed of light is c.

The binding energy per nucleon for the parent nucleus is

E_{1}

and that for the daughter nuclei is

E_{2}

. Then

(A) $E_{2} = 2 E_{1}$
(B) $E_{1} > E_{2}$
(C) $E_{2} > E_{1}$ ✓ Correct
(D) $E_{1} = 2 E_{2}$

Answer: (C)

Solution

When a nucleus splits into fragments, the products are more tightly bound than the parent, and that extra binding is exactly what lets the decay release energy. Greater stability means a larger binding energy per nucleon, so the daughter nuclei have a higher binding energy per nucleon than the parent. Hence

E_{2} > E_{1}

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option

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