Q.Obtain the binding energy (in MeV) of a nitrogen nucleus $^{14}_{7}\text{N}$, given $m\left(^{14}_{7}\text{N}\right) = 14.00307\ \text{u}$.

Concept: Mass Energy Equivalence – the binding energy is the energy equivalent of the mass defect, using $1\ \text{u} = 931.5\ \text{MeV}/c^2$.

Step 1 – Find the total mass of constituents.

A $^{14}_{7}\text{N}$ nucleus has 7 protons and 7 neutrons.

Mass of 7 protons: $7 \times 1.007825\ \text{u} = 7.054775\ \text{u}$

Mass of 7 neutrons: $7 \times 1.008665\ \text{u} = 7.060655\ \text{u}$

Total mass of nucleons: $7.054775 + 7.060655 = 14.11543\ \text{u}$

Step 2 – Compute the mass defect.

$\Delta m = \text{(mass of nucleons)} - \text{(actual nuclear mass)}$

$\Delta m = 14.11543 - 14.00307 = 0.11236\ \text{u}$

Step 3 – Convert to energy.

Binding energy $E_b = \Delta m \times 931.5\ \text{MeV/u}$

$E_b = 0.11236 \times 931.5 \approx 104.66\ \text{MeV}$

Nitrogen $^{14}_{7}\text{N}$ has $Z=7$ protons and $N = 14-7 = 7$ neutrons. First find the mass defect using $m(^{1}_{1}\text{H})=1.007825\ \text{u}$, $m_n = 1.008665\ \text{u}$ and the given $m(^{14}_{7}\text{N}) = 14.00307\ \text{u}$:

The binding energy is the energy equivalent of this mass defect, using $1\ \text{u} = 931.5\ \text{MeV}/c^2$:

Method: Mass Defect → Binding Energy via Einstein's Mass-Energy Equivalence

The binding energy of a nucleus is the energy equivalent of the mass defect — the difference between the sum of masses of its individual nucleons and the actual nuclear mass. The method uses Einstein's relation $E = \Delta m c^2$, converting atomic mass units (u) directly to MeV using the standard conversion factor.

Step 1: Identify the composition of the nucleus

For $^{14}_{7}\text{N}$:

• Atomic number $Z = 7$ → 7 protons
• Mass number $A = 14$ → number of neutrons = $A - Z = 14 - 7 = 7$ neutrons

So the nucleus contains 7 protons and 7 neutrons.

Step 2: Write the mass of the individual constituents (in u)

The given nuclear mass $14.00307\ \text{u}$ is an atomic mass (it includes the atom's electrons). To make the electron masses cancel automatically, compare against $Z$ hydrogen ATOMS (each carrying its own electron) rather than bare protons:

• Mass of a hydrogen atom, $m_H = 1.007825\ \text{u}$ (proton + its electron)
• Mass of a neutron, $m_n = 1.008665\ \text{u}$

Step 3: Calculate the total mass of the separated constituents

Step 4: Find the mass defect

Mass defect $\Delta m$ = (mass of constituents) − (actual atomic mass)

Step 5: Convert mass defect to energy

Use the standard conversion: $1\ \text{u} = 931.5\ \text{MeV}/c^2$

Final answer:

Binding energy of $^{14}_{7}\text{N}$ ≈ 104.66 MeV

Common Mistakes in Binding Energy Problems (Mass-Energy Equivalence)

Students lose marks on this exact type of question in predictable ways. Here are the most frequent errors and how to fix them.

Mistake 1: Forgetting to account for the mass of electrons

The given mass $m(^{14}_{7}\text{N}) = 14.00307\ \text{u}$ is the atomic mass — it includes the mass of 7 electrons. But when you calculate the mass defect, you need the nuclear mass of nitrogen, not the atomic mass.

What students do wrong: They directly subtract the given mass from the sum of proton and neutron masses, forgetting that the proton mass given in data tables is also the mass of a hydrogen atom (proton + electron).

How to avoid: Always use atomic mass units consistently. The mass of a hydrogen atom $m(^1_1\text{H}) = 1.007825\ \text{u}$ already includes one electron. So for a nucleus with $Z$ protons, the total mass of $Z$ hydrogen atoms automatically accounts for $Z$ electrons — matching the $Z$ electrons already included in the atomic mass of the nucleus.

For $^{14}_7\text{N}$:

• $Z = 7$, $A = 14$
• $m(^1_1\text{H}) = 1.007825\ \text{u}$
• $m_n = 1.008665\ \text{u}$
• $m(^{14}_7\text{N}) = 14.00307\ \text{u}$

So:

Mistake 2: Using the wrong conversion factor from u to MeV

The standard conversion is $1\ \text{u} = 931.5\ \text{MeV}/c^2$. Some students use 931 or 931.5 MeV — both are accepted in most boards, but be consistent with what your textbook uses.

What students do wrong: They forget the $c^2$ and treat the mass defect as if it's already in energy units, or they use the wrong conversion factor entirely.

How to avoid: Write the conversion explicitly:

Mistake 3: Arithmetic errors in the mass defect calculation

This is the most common — and most frustrating — mistake. The numbers are close together, and a small slip changes the answer completely.

What students do wrong: They mis-add or mis-subtract the 7-digit numbers, or they round too early.

How to avoid: Do the calculation step by step and keep at least 5 decimal places until the final answer.

Let's do it properly:

• $7 \times 1.007825 = 7.054775$
• $7 \times 1.008665 = 7.060655$
• Sum = $7.054775 + 7.060655 = 14.115430$
• Subtract given mass: $14.115430 - 14.003070 = 0.112360\ \text{u}$

Mistake 4: Forgetting to multiply by $c^2$ or misplacing the conversion

Some students compute $\Delta m$ correctly but then write $E = \Delta m \times c^2$ without converting units, getting a meaningless number.

How to avoid: Remember that $1\ \text{u} = 931.5\ \text{MeV}/c^2$, so:

Mistake 5: Reporting the wrong number of significant figures

The given mass is $14.00307\ \text{u}$ (6 significant figures). The proton and neutron masses are typically given to 6 or 7 figures. Your final answer should reflect this precision.

What students do wrong: They round to 2 or 3 significant figures, or they report 104.66 MeV when the data only justifies 104.7 MeV.

How to avoid: Keep intermediate calculations to 5-6 decimal places, then round the final energy to match the least precise input. Here, 104.7 MeV is appropriate.

Mistake 6: Confusing binding energy per nucleon with total binding energy

The question asks for binding energy — that's the total. Some students divide by 14 and report the per-nucleon value instead.

How to avoid: Read the question carefully. If it says "binding energy" without "per nucleon," give the total. If you want to be safe, you can state both, but clearly label which is which.

Quick Checklist to Avoid All These Mistakes

1. Use hydrogen atom mass (not proton mass) for the protons
2. Subtract the atomic mass of the nucleus (which includes electrons)
3. Keep 5-6 decimal places in $\Delta m$
4. Multiply by 931.5 to get MeV
5. Round final answer appropriately
6. Check if the question wants total or per-nucleon binding energy