ΔE = Δm × c². Calculate binding energy per nucleon for He-4.

Nuclear Binding Energy — Mass Defect Calculation

Question

The atomic mass of helium-4 ( $^4_2\text{He}$ ) is 4.0026 u. Given that the mass of a proton is 1.0073 u and the mass of a neutron is 1.0087 u, calculate:

The mass defect ( $\Delta m$ )
The binding energy of the nucleus
The binding energy per nucleon

Use: 1 u = 931.5 MeV/c²

Solution — Step by Step

Helium-4 has atomic number $Z = 2$ , so it has 2 protons and 2 neutrons (since mass number $A = 4$ , neutrons $= A - Z = 2$ ).

This is the most common stumbling block — students confuse atomic number with neutron count. Always subtract $Z$ from $A$ .

Sum the masses of all 2 protons and 2 neutrons as if they were completely separate:

m_{\text{free}} = 2m_p + 2m_n = 2(1.0073) + 2(1.0087)

m_{\text{free}} = 2.0146 + 2.0174 = 4.0320 \text{ u}

The actual measured mass of He-4 is less than the sum of its parts. Nature “paid” some mass to bind the nucleus together. This missing mass is $\Delta m$ :

\Delta m = m_{\text{free}} - m_{\text{nucleus}} = 4.0320 - 4.0026

\boxed{\Delta m = 0.0294 \text{ u}}

Einstein’s mass-energy equivalence does the conversion. Since 1 u = 931.5 MeV:

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

E_b = 0.0294 \times 931.5 = \boxed{27.39 \text{ MeV}}

Divide total binding energy by the total number of nucleons ( $A = 4$ ):

\frac{E_b}{A} = \frac{27.39}{4} = \boxed{6.85 \text{ MeV/nucleon}}

This value is typical for light nuclei. For reference, Fe-56 sits at ~8.8 MeV/nucleon — the most stable nucleus in the periodic table.

Why This Works

When nucleons bind together to form a nucleus, the strong nuclear force pulls them into a lower energy state. To reach that lower state, the system must release energy — and by $E = mc^2$ , releasing energy means losing mass. The “missing” mass ( $\Delta m$ ) is exactly the energy the nucleus released when it formed.

This is why splitting heavy nuclei (fission) or fusing light ones (fusion) both release energy. In both cases, the products have higher binding energy per nucleon than the reactants. The mass difference escapes as kinetic energy and radiation.

Binding energy per nucleon is the real figure of merit. He-4’s ~7 MeV/nucleon is why alpha particles are so stable — they’re released intact in radioactive decay rather than breaking apart.

Alternative Method — Using Atomic Masses Directly

In NEET and JEE problems, you’re sometimes given atomic masses (which include electrons) instead of nuclear masses.

For He-4, atomic mass includes 2 electrons. So we use:

\Delta m = [Zm_H + Nm_n] - M_{\text{atom}}

where $m_H = 1.00783$ u (atomic mass of hydrogen, which includes one electron).

\Delta m = [2(1.00783) + 2(1.00867)] - 4.00260

= 4.03300 - 4.00260 = 0.03040 \text{ u}

The electron masses cancel out automatically when you use atomic hydrogen mass ( $m_H$ ) instead of proton mass ( $m_p$ ). This is the cleaner method when the question gives you atomic masses — you don’t need to subtract electron masses separately.

The small difference from our earlier answer (~0.03040 vs 0.02940 u) comes from the slightly different input values. Both approaches are valid — use whichever mass values the question provides.

Common Mistake

Using nuclear mass = atomic mass directly. Atomic mass includes the electrons. He-4 has 2 electrons, each with mass ~0.000549 u. If the question gives atomic mass and you use proton mass (not $m_H$ ), you’ll double-count the electron masses and get a wrong $\Delta m$ .

Always check: is the given mass the nuclear mass or atomic mass? If it’s atomic mass (which it usually is in data tables), pair it with $m_H = 1.00783$ u for hydrogen, not the bare proton mass 1.00727 u.

This exact distinction appeared in NEET 2024 — the question gave atomic masses, and students who used $m_p = 1.0073$ u instead of $m_H = 1.00783$ u got a slightly off answer and lost the mark.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Atomic Masses Directly

Common Mistake

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