Question

Which of the following has the most stable nucleus? (Binding energy is given in the table)

$\text{Nucleus}$	$\text{Binding Energy}(B.E)$
${}_2{\text{He}}^4$	$28\text{MeV}$
${}_3{\text{Li}}^7$	$42\text{MeV}$
${}_6{\text{C}}^{12}$	$53\text{MeV}$

• A. Helium
• B. Lithium
• C. Carbon
• D. All have equal stability

Answer 1

The correct option is A Helium

Stability of a nucleus is decided by binding energy per nucleon.

Greater the $B.E$ per nucleon, greater will be stability.

$\text{stability}\propto \frac{B.E}{A}$

Nucleus	$B.E$	$\frac{B.E}{A}$
${}_2{\text{He}}^4$	$28\text{MeV}$	$7\text{MeV}$
${}_3{\text{Li}}^7$	$42\text{MeV}$	$6\text{MeV}$
${}_6{\text{C}}^{12}$	$53\text{MeV}$	$4.4\text{MeV}$

As $\frac{B.E}{A}$ is higher for ${}^4{\text{He}}_2$,

$\Rightarrow {}^4{\text{He}}_2$ is the most stable nucleus among the given nuclei.

Why this Question? Greater the $B.E$ per nucleon, greater will be stability. $\text{stability}\propto \frac{B.E}{A}$