Q.Obtain the binding energy of the nuclei $^{56}_{26}\text{Fe}$ and $^{209}_{83}\text{Bi}$ in units of MeV from the following data:
$m\left(^{56}_{26}\text{Fe}\right) = 55.934939\ \text{u}$,
$m\left(^{209}_{83}\text{Bi}\right) = 208.980388\ \text{u}$.

The binding energy is the energy equivalent of the mass defect, $B = [Z\,m_H + (A-Z)m_n - M]\,c^2$, using atomic masses with $1\ \text{u} = 931.5\ \text{MeV}/c^2$, $m_H = 1.007825\ \text{u}$, $m_n = 1.008665\ \text{u}$.

$^{56}_{26}\text{Fe}$ ($Z=26$, $N=30$):

$^{209}_{83}\text{Bi}$ ($Z=83$, $N=126$): …

Principle

The measured mass of a nucleus is less than the total mass of its separate protons and neutrons. This missing mass — the mass defect $\Delta m$ — is the mass equivalent of the energy that binds the nucleons together. The binding energy is

Because the tabulated nuclear masses are atomic masses (they include the electrons), we use the hydrogen-atom mass $m_H = 1.007825\ \text{u}$ for each proton so the electron masses cancel. With $m_n = 1.008665\ \text{u}$ and $1\ \text{u} = 931.5\ \text{MeV}/c^2$, multiplying $\Delta m$ (in u) by $931.5$ gives $B$ in MeV directly.

$^{56}_{26}\text{Fe}$ ($Z = 26$, $N = 30$)

Mass of the constituents:

Mass defect:

Binding energy:

$^{209}_{83}\text{Bi}$ ($Z = 83$, $N = 126$)

Mass of the constituents: …

Method: Mass Defect → Binding Energy (Einstein’s mass-energy equivalence)

The binding energy of a nucleus is the energy required to separate it into its individual protons and neutrons. It is found by first calculating the mass defect — the difference between the mass of the separated nucleons and the actual nuclear mass — then converting that mass difference into energy using $E = \Delta m c^2$, with the conversion factor $1\ \text{u} = 931.5\ \text{MeV}/c^2$.

Steps for each nucleus:

1. Identify the number of protons ($Z$) and neutrons ($N$).

   For $^{56}_{26}\text{Fe}$: $Z = 26$, $N = 56 - 26 = 30$.

   For $^{209}_{83}\text{Bi}$: $Z = 83$, $N = 209 - 83 = 126$.

2. Write the mass of the separated constituents — atomic mass throughout.

   The given nuclear masses are atomic masses (electrons included), so pair them with the hydrogen ATOM mass $m_H = 1.007825\ \text{u}$ (not the bare proton mass $m_p=1.007276\ \text{u}$ — that would silently drop $Z$ electron masses) and the neutron mass $m_n = 1.008665\ \text{u}$.

   Total mass of constituents = $Z \cdot m_H + N \cdot m_n$.

3. Compute the mass defect $\Delta m$.

   $\Delta m = (\text{mass of constituents}) - (\text{actual atomic mass})$.

4. Convert $\Delta m$ to energy.

   Binding energy $B = \Delta m \times 931.5\ \text{MeV}$.

For $^{56}_{26}\text{Fe}$:

• Constituent mass = $26 \times 1.007825 + 30 \times 1.008665$ $= 26.203450 + 30.259950 = 56.463400\ \text{u}$.
• $\Delta m = 56.463400 - 55.934939 = 0.528461\ \text{u}$.
• $B = 0.528461 \times 931.5 \approx 492.3\ \text{MeV}$.

For $^{209}_{83}\text{Bi}$:

• Constituent mass = $83 \times 1.007825 + 126 \times 1.008665$ …

Common Mistakes in Nuclear Binding Energy Calculations

Students often lose marks on this exact type of problem. Here are the most frequent errors and how to avoid each.

Mistake 1: Forgetting to account for electrons in atomic masses

The given masses are atomic masses (include electrons), but the binding energy formula uses nuclear masses. The mass of Z electrons is already in the atomic mass — if you subtract only the proton mass from the atomic mass, you double-count the electrons.

How to avoid: Use the standard shortcut: treat the hydrogen atom mass $m(^1\text{H}) = 1.007825\ \text{u}$ as the combined mass of one proton and one electron. Then:

This automatically handles the electron mass correctly.

Mistake 2: Using wrong number of neutrons

For $^{56}_{26}\text{Fe}$, $N = 56 - 26 = 30$. For $^{209}_{83}\text{Bi}$, $N = 209 - 83 = 126$. Students sometimes confuse $A$ and $Z$ or subtract incorrectly.

How to avoid: Write $N = A - Z$ clearly before plugging in. Double-check: the mass number $A$ is the superscript, atomic number $Z$ is the subscript.

Mistake 3: Using proton mass instead of hydrogen mass

If you use $m_p = 1.007276\ \text{u}$ directly, you must also add $Z \cdot m_e$ separately. Most students forget the electron term.

How to avoid: Stick to the hydrogen-atom shortcut above. It's the standard method in NCERT and all board exams.

Mistake 4: Arithmetic errors in the mass defect

The mass defect $\Delta m$ is small — a few tenths of a u. Students often misalign decimal places when subtracting.

For Fe:

• $Z \cdot m(^1\text{H}) = 26 \times 1.007825 = 26.20345\ \text{u}$
• $N \cdot m_n = 30 \times 1.008665 = 30.25995\ \text{u}$
• Sum = $56.46340\ \text{u}$
• Subtract $m(^{56}\text{Fe}) = 55.934939\ \text{u}$
• $\Delta m = 0.528461\ \text{u}$

How to avoid: Write each term to 6 decimal places and subtract carefully. Use column subtraction, not mental math.

Mistake 5: Using 931 MeV/u incorrectly …