Calculate binding energy per nucleon of He-4 given atomic masses

Question

Calculate the binding energy per nucleon of helium-4 ( ${}^4_2\text{He}$ ) given:

Mass of proton $m_p = 1.007825$ u
Mass of neutron $m_n = 1.008665$ u
Atomic mass of He-4 $M = 4.002602$ u
$1$ u $= 931.5$ MeV/ $c^2$

Solution — Step by Step

Helium-4 has:

$Z = 2$ protons
$A - Z = 4 - 2 = 2$ neutrons

So the nucleus contains 2 protons and 2 neutrons (this is the alpha particle).

If the protons and neutrons were completely free (not bound):

M_{free} = Z \cdot m_p + (A-Z) \cdot m_n

M_{free} = 2 \times 1.007825 + 2 \times 1.008665

M_{free} = 2.015650 + 2.017330 = 4.032980 \text{ u}

The actual atomic mass of He-4 is $4.002602$ u. The mass defect:

\Delta m = M_{free} - M_{He-4}

\Delta m = 4.032980 - 4.002602 = 0.030378 \text{ u}

This is the “missing” mass — it was converted to binding energy when the nucleus formed.

Using $1$ u $= 931.5$ MeV:

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

E_b = 0.030378 \times 931.5 = 28.3 \text{ MeV}

\frac{E_b}{A} = \frac{28.3 \text{ MeV}}{4} = 7.07 \text{ MeV per nucleon}

\boxed{\frac{E_b}{A} \approx 7.07 \text{ MeV/nucleon}}

This is fairly high — the helium-4 nucleus (alpha particle) is exceptionally stable for its size, which is why alpha particles are emitted in radioactive decay rather than individual protons or neutrons.

Why This Works

The mass defect directly measures how much energy was released when the nucleus formed — or equivalently, how much energy you’d need to supply to break it apart into free protons and neutrons.

The conversion factor $1$ u $= 931.5$ MeV comes from $E = mc^2$ with $m = 1$ u $= 1.66054 \times 10^{-27}$ kg:

E = (1.66054 \times 10^{-27})(3 \times 10^8)^2 = 1.4924 \times 10^{-10} \text{ J} = 931.5 \text{ MeV}

Binding energy per nucleon is the most useful way to compare nuclear stability across different elements — it removes the trivial effect of a larger nucleus having more total binding energy.

Alternative Method

A shortcut for BE/A calculations in exams: once you know $\Delta m$ in atomic mass units, just multiply by 931.5 and divide by $A$ . You don’t need to find total $E_b$ as a separate step:

\frac{E_b}{A} = \frac{\Delta m \times 931.5}{A} = \frac{0.030378 \times 931.5}{4} \approx 7.07 \text{ MeV/nucleon}

Common Mistake

Students sometimes use the nuclear mass instead of the atomic mass in calculations, or confuse the two. The problem usually gives atomic masses (which include electron masses). When calculating mass defect using atomic masses, you use:

$\Delta m = Z \cdot m_H + N \cdot m_n - M_{atom}$

where $m_H = 1.007825$ u is the atomic mass of hydrogen (proton + electron), not just the proton mass. This way, the electron masses cancel automatically. If you use bare proton mass with atomic masses, you introduce a small error per electron.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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