Binding energy — mass defect, BE per nucleon curve: Question bank

Before we test you, one 10-second refresher on the shape everything hangs on:

The binding-energy-per-nucleon curve plots $\overline{E_{B}}$ (vertical, in MeV) against mass number $A$ = total nucleons (horizontal).
It rises fast on the left (light nuclei), peaks near $A \approx 56$ (iron region, $\approx 8.8$ MeV), then falls slowly on the right (heavy nuclei).
Higher up = more tightly glued = more stable. Reactions that move nucleons toward the peak release energy.

Keep that mental picture ready. Every trap below is really a question about where you sit on this curve or which way mass and energy point.

True or false — justify

A nucleus is heavier than the sum of its free protons and neutrons.

False. It is lighter — some mass became the binding (glue) energy that left the system when the nucleons clumped, so by

E = m c^{2}

the nucleus weighs less. See Mass–energy equivalence ($E=mc^2$).

Binding energy is energy stored inside the nucleus that you could tap by keeping it intact.

False. Binding energy already left the system when the nucleus formed; it is the energy you must pay back in to break the nucleus apart, not a reserve you extract by doing nothing.

A larger total binding energy

E_{B}

always means a more stable nucleus.

False. Total

E_{B}

climbs with

A

because more nucleons make more bonds;

^{238}

U has huge total

E_{B}

but a low

\overline{E_{B}} \approx 7.6

MeV, so it can still fission. Stability is judged by

\overline{E_{B}}

, the per-nucleon value.

^{4}

He has a higher BE per nucleon than its light neighbours like

^{3}

H or

^{6}

Li.

True.

^{4}

He is a local peak on the curve — doubly closed shells make it exceptionally tightly glued (

\approx 7.1

MeV/nucleon), which is why it appears as the sharp spike on the left slope.

Since fusion releases energy, fusing any two nuclei releases energy.

False. Only fusion that moves the product upward on the curve releases energy — that means light nuclei below the peak. Fusing two nuclei already past iron would move down the curve and cost energy. See Nuclear fusion.

Iron (

^{56}

Fe) can release energy by either fission or fusion.

False. Iron sits at the peak, so any move — splitting or joining — goes downhill to lower

\overline{E_{B}}

, which absorbs energy. Iron is the stellar "ash," the end of energy-releasing reactions.

The mass defect and the binding energy are two different physical quantities.

Half-true — they are the same fact in two units.

E_{B} = Δ m c^{2}

; the "missing mass" and the "glue energy" are just the mass and energy descriptions of one identical thing.

You should always use nuclear masses in

Δ m = Z m_{p} + N m_{n} - M

False in practice. Tables list atomic masses (nucleus +

Z

electrons). If you use

m_{^{1} H}

for protons and the atomic mass for

M

, the

Z

electron masses cancel automatically — cleaner and correct.

Spot the error

"Adding energy to a nucleus makes it heavier, so a strongly bound nucleus is a heavy nucleus."

The error is direction: binding energy left the nucleus. A strongly bound nucleus is lighter (bigger mass defect), and "heavy" (

A

, mass number) is unrelated to "tightly bound" (

\overline{E_{B}}

Δ m = Z m_{p} + N m_{n} - M_{nucleus}

, and I'll plug in the atomic mass from the table for

M_{nucleus}

The error is mixing conventions:

m_{p}

is a bare proton but the table mass includes

Z

electrons. Either use all-nuclear or all-atomic masses; the standard fix uses

m_{^{1} H}

so electrons cancel.

"BE per nucleon rises forever as nuclei get bigger, because more nucleons make more bonds."

The error ignores Coulomb repulsion. Beyond

A \approx 56

, long-range proton–proton repulsion (growing as

Z^{2}

) outpaces the short-range Strong nuclear force, so

\overline{E_{B}}

falls — the curve turns over.

"Fission releases energy because a big nucleus breaks the strong force bonds."

Backwards. Breaking bonds costs energy. Fission releases energy because the products are more tightly bound per nucleon (higher on the curve); the net gain in binding is what comes out. See Nuclear fission.

"In D+T→He+n the neutron carries no energy, so all

17.6

MeV goes to the helium."

The error is skipping momentum conservation: the light neutron actually carries away most of the kinetic energy. The

17.6

MeV is the total

Q

-value, shared, not the helium's share. See Q-value of nuclear reactions.

1 u = 931.5

MeV, so mass and energy have the same units."

Sloppy: it should read

1 u \cdot c^{2} = 931.5

MeV. Mass and energy differ by a factor

c^{2}

; dropping the

c^{2}

hides the unit conversion.

"The strong force reaches across the whole nucleus, so a giant nucleus is glued everywhere equally."

The error is the range. The strong force is short-range (nearest-neighbour only), so a nucleon deep inside feels only its neighbours, while every proton repels every other proton across the whole nucleus — the imbalance that ruins heavy-nucleus stability.

Why questions

Why do we divide by

A

instead of comparing total binding energies?

Total

E_{B}

unfairly rewards big nuclei (more particles = more bonds). Dividing by

A

asks "how hard to pull out one average nucleon?" — a fair, size-independent stability comparison.

Why does the left slope of the curve rise so steeply?

In a tiny nucleus, adding one nucleon suddenly gives it many new attractive neighbours, so binding jumps a lot per particle. As the nucleus grows, new nucleons on the surface gain fewer new neighbours, so the payoff flattens.

Why is the mass defect always positive for a stable bound nucleus?

A bound state sits in an energy well, lower in energy than the free nucleons; lower energy means lower mass, so the separated pieces always outweigh the whole — a positive

Δ m

Why do both fusion and fission release energy despite being opposite processes?

Both move nucleons toward the peak of the curve. Light nuclei climb up by fusing; heavy nuclei climb up by splitting. The rise in

\overline{E_{B}}

is the energy released in either direction. See Nuclear fusion and Nuclear fission.

Why does the curve peak around iron and not, say, lead?

Iron is the sweet spot where short-range attraction is nearly saturated but Coulomb repulsion hasn't yet dominated. Past iron, the

Z^{2}

repulsion wins and

\overline{E_{B}}

declines, so lead is below the peak.

Why can we ignore electron masses if we use atomic masses on both sides?

Atomic masses each carry their electrons:

Z

electrons in the

Z

hydrogen atoms match the

Z

electrons in the neutral product atom, so they cancel exactly (electron binding energies are tiny and neglected).

Why is

^{4}

He's BE per nucleon high even though it is a light nucleus?

Its neutrons and protons fill closed shells ("doubly magic"), an especially stable arrangement, producing a sharp local peak far to the left of iron. It's an exception to the smooth-curve trend.

Edge cases

What is the binding energy of a single free proton (hydrogen-1 nucleus)?

It's zero — there is nothing holding together, no partner nucleon, so no mass defect and no glue energy. It sits at

\overline{E_{B}} = 0

, the very start of the curve.

For a lone neutron, is there a binding energy?

No — a single nucleon has no bonds, so

E_{B} = 0

. (A free neutron is unstable and decays, but that's beta decay, unrelated to nuclear binding.)

At the exact peak of the curve, what energy does a reaction release?

Essentially zero net energy for any small rearrangement, because moving either way (fission or fusion) leaves the peak and lowers

\overline{E_{B}}

, so those directions cost energy — the peak is the energy floor for the products.

If a "nucleus" had positive mass defect defined the other way (nucleus heavier than parts), what would it mean?

It would be unbound — the pieces would fly apart on their own, since the assembled state is higher in energy than the free nucleons. Real stable nuclei never do this.

What does

Δ m = 0

physically correspond to?

A completely unbound collection: the whole equals the sum of parts, no glue, no released energy — the borderline case with zero binding energy per nucleon.

Does the very-light region ever dip rather than rise smoothly?

Yes — the curve is jagged for tiny

A

^{4}

He spikes high, but

^{6}

Li and

^{7}

Li dip below it, and

^{8}

Be is unstable. The smooth "rise" is only the overall trend, not every point.

Connections

Mass–energy equivalence ($E=mc^2$)
Nuclear fission
Nuclear fusion
Strong nuclear force
Stability of nuclei & N-Z curve
Q-value of nuclear reactions
Atomic mass unit (u)