Q.Determine the binding energy per nucleon of the americium isotope $_{95}^{244}$Am, given the mass of $_{95}^{244}$Am to be 244.06428 u.

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)

This 'missing' mass has been converted into the binding energy (via E = delta_m*c^2) that holds the nucleons together inside the nucleus - it is the energy that would need to be supplied to pull the nucleus apart into its separate, free nucleons.

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

Binding energy released: $E = \Delta m\, c^2 = 0.004 \times (3\times10^8)^2 = 0.004 \times 9\times10^{16} = 3.6\times10^{14}\,\text{J}$.

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.

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

Because $Q>0$ (energy is emitted), $\Delta m$ must be positive, which requires

$(m_1 + m_2) > m.$

The fused nucleus is more tightly bound (higher binding energy per nucleon), so it is lighter than the two separate nuclei.

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:

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

This is the energy that would need to be supplied to break the nucleus apart into its free, separated nucleons — it is numerically and physically equivalent to the mass defect, not to the mass of a proton, neutron, or the nucleus itself.

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:

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

Checking the options: (a) omits $C^2$ and has the sign reversed; (b) wrongly uses $A$ neutrons instead of $(A-Z)$ neutrons; (d) has the mass-defect terms in the wrong (reversed) order, which would make $B\cdot E$ negative. Only option (c) is dimensionally and physically correct.

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 $=15.9994\,u$.

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

$Q = \Delta m \times 931.5\ \text{MeV/u} = 0.0110\times931.5 \approx 10.24\ \text{MeV}$

This is the total energy released per oxygen nucleus formed.

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

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

where $m_p$, $m_n$ are the free proton and neutron masses and $M_{nucleus}$ is the actual (measured) mass of the nucleus. This mass defect is always positive - the bound nucleus has less mass than its separated constituents - because, by Einstein's mass-energy relation $E=mc^2$, this 'missing' mass $\Delta m\, c^2$ is exactly the binding energy that was released when the nucleons came together (and would have to be supplied to pull the nucleus apart again).

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:

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

This $E_b$ is called the binding energy of the nucleus — it is the energy that was released on formation, and equivalently it is the minimum energy that must be supplied to completely separate the nucleus back into its individual free protons and neutrons at rest, infinitely far apart.