Q.Why is a nucleus more stable if it has large value of mass defect? Explain.

A bound nucleus weighs slightly LESS than the sum of its free, separate nucleons - that missing mass is the mass defect.

If you add up the masses of Z free protons and (A-Z) free neutrons that would make up a nucleus of mass number A, this sum is always slightly GREATER than the actual measured mass of the bound nucleus. This difference is called the mass defect:

delta_m = [Z*m_p + (A-Z)*m_n] - M(nucleus)

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Using Einstein's mass-energy relation $E = \Delta m\, c^2$ with $\Delta m = 0.4\%$ of 1 kg gives $3.6\times10^{14}\,\text{J}$.

Mass defect $\Delta m = 0.4\%\ \text{of}\ 1\,\text{kg} = 0.004\,\text{kg}$.

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Energy is released only if some mass disappears; the product mass $m$ is less than $m_1+m_2$, so $(m_1+m_2) > m$ — option (B).

Concept — mass–energy equivalence. In fusion, two light nuclei combine. If the process releases energy $Q$, that energy comes from a loss of mass (the mass defect $\Delta m$) through Einstein's relation $E = \Delta m\,c^2$.

Reasoning.

$\Delta m = (m_1 + m_2) - m > 0 \quad\Rightarrow\quad Q = \Delta m\,c^2 > 0.$ …

A nucleus's mass is always slightly less than the sum of the masses of its separate nucleons; this missing mass, the mass defect, is exactly equivalent (via $E=mc^2$) to the binding energy released when the nucleus formed.

Mass defect: $\Delta m = [Zm_p + (A-Z)m_n] - M_{nucleus}$

Binding energy is the energy equivalent of this mass defect:

$E_b = \Delta m\, c^2$

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Binding energy equals the mass defect (sum of the masses of the free constituent nucleons minus the actual nuclear mass) multiplied by $C^2$; matching this to the options gives option (c).

Mass defect

A nucleus $_ZX^A$ contains $Z$ protons and $(A-Z)$ neutrons. If these nucleons existed freely (unbound), their total mass would be $ZM_p + (A-Z)M_n$. The actual measured mass of the bound nucleus, $M(A,Z)$, is less than this because some mass is converted into the energy that binds the nucleus together. This mass difference is the mass defect:

$\Delta m = ZM_p + (A-Z)M_n - M(A,Z)$

Binding energy

By Einstein's mass-energy relation, this "missing" mass corresponds to the binding energy:

$B\cdot E = \Delta m\, C^2 = \big[ZM_p + (A-Z)M_n - M(A,Z)\big]C^2$

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4 He nuclei fuse into 1 O nucleus; the mass defect times $931.5\,$MeV/u gives the energy released per O nucleus.

Since $4\times4=16$, the fusion of 4 helium (He-4) nuclei into one oxygen (O-16) nucleus is the reaction implied:

$4\,^4_2\text{He} \to\ ^{16}_{8}\text{O}$

Mass of 4 He nuclei $= 4\times4.0026 = 16.0104\,u$. Mass of one O nucleus $=15.9994\,u$.

$\Delta m = 16.0104 - 15.9994 = 0.0110\,u$ …

A nucleus always weighs slightly less than the sum of its separate constituent nucleons; this 'missing' mass is the mass defect, and by E = mc^2 it corresponds to the nuclear binding energy.

For a nucleus $^A_Z X$ made of Z protons and (A-Z) neutrons, the mass defect is defined as

$\Delta m = \big[Z m_p + (A-Z)m_n\big] - M_{nucleus}$

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Nuclear binding energy = $\Delta m\,c^2$, the energy equivalent of the 'missing' mass when free nucleons bind together into a nucleus; it is also the energy needed to pull the nucleus apart again.

When $Z$ protons and $N$ neutrons combine to form a nucleus, the mass of the resulting nucleus $M_{nucleus}$ is always slightly LESS than the sum of the masses of the free, separated nucleons:

$\Delta m = [Zm_p + Nm_n] - M_{nucleus}$

This mass difference $\Delta m$ (the 'mass defect') is converted to energy and released when the nucleus is formed, in accordance with Einstein's mass–energy relation:

$E_b = \Delta m\,c^2$ …