Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.

Q: Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.

**The plot** Potential energy U is plotted on the y-axis against nucleon separation r on the x-axis: - For $r > r_0$ ($\approx 0.8 \text{ fm}$), U is negative and rises towards zero as r increases — the force is **attractive**. - U reaches a **minimum of about $-100 \text{ MeV}$ at $r = r_0 \approx 0.8 \text{ fm}$** (stable separation). - For $r 0.8 \text{ fm}$ and repulsive for $r < 0.8 \text{ fm}$**, with equilibrium (minimum PE) at $r_0 \approx 0.8 \text{ fm}$. 2. The force is **very short-ranged**: $U \rightarrow 0$ within a few femtometres, so nucleons interact only with their nearest neighbours (saturation of nuclear force).

Q: What does the minimum of the potential-energy curve represent?

The minimum (about $-100 \text{ MeV}$) at $r_0 \approx 0.8 \text{ fm}$ marks the stable equilibrium separation of two nucleons, where the net force between them is zero. Nucleons naturally settle at this spacing inside a nucleus.

Q: How does the graph show the nuclear force is short-ranged?

The potential energy rapidly approaches zero as the separation exceeds a few femtometres. Since force is the negative gradient of U, a flat curve means negligible force — so nucleons feel the strong force only over very short distances, with each nucleon interacting mainly with its nearest neighbours.

Nuclei

Nuclear force — potential energy of nucleons

Official Answer

The plot

Potential energy U is plotted on the y-axis against nucleon separation r on the x-axis:

For $r > r_0$ ( $\approx 0.8 \text{ fm}$ ), U is negative and rises towards zero as r increases — the force is attractive.
U reaches a minimum of about $-100 \text{ MeV}$ at $r = r_0 \approx 0.8 \text{ fm}$ (stable separation).
For $r < r_0$ , U becomes positive and steeply rising — the force is strongly repulsive.

Two conclusions

The nuclear force is attractive for $r > 0.8 \text{ fm}$ and repulsive for $r < 0.8 \text{ fm}$ , with equilibrium (minimum PE) at $r_0 \approx 0.8 \text{ fm}$ .
The force is very short-ranged: $U \rightarrow 0$ within a few femtometres, so nucleons interact only with their nearest neighbours (saturation of nuclear force).

potential energy nucleonsnuclear forcer0 0.8 fmattractive and repulsiveminimum potential energyshort range forcesaturation-100 MeV

Marking Scheme

11 mark: correct U-vs-r plot showing a minimum (~ $-100 \text{ MeV}$ ) at $r_0 \approx 0.8 \text{ fm}$ , a repulsive positive region for $r < r_0$ , and $U \rightarrow 0$ for large r.
20.5 mark each (×2): two valid conclusions — (i) attractive for $r > 0.8 \text{ fm}$ , repulsive for $r < 0.8 \text{ fm}$ ; (ii) nuclear force is short-ranged / saturates.

Hint

The curve dips to a minimum near $r_0 \approx 0.8 \text{ fm}$ — attractive beyond it, repulsive within it, and flat (zero) beyond a few femtometres.

Quick Oral Answer

The potential-energy curve dips to a minimum of about minus 100 MeV at a separation of roughly 0.8 femtometres — the force is attractive beyond this distance, strongly repulsive below it, and dies out within a few femtometres, showing it is short-ranged.

Analysis & Explanation

Concept

The potential-energy curve encodes the nuclear force through $F = -dU/dr$ . Where the curve slopes upward with increasing r ( $r > r_0$ ), the force is attractive; where it plunges as r decreases below r₀, the force is repulsive — this repulsive core stops the nucleus from collapsing.

Reading the graph

The minimum at $r_0 \approx 0.8 \text{ fm}$ is the equilibrium spacing of nucleons, where net force is zero.
The depth (~100 MeV) reflects how strongly nucleons are bound — far stronger than electromagnetic binding of electrons (eV scale).

Exam trap

Do not draw the curve like the Coulomb ( $1/r$ ) curve; the nuclear curve has a minimum and a repulsive core, unlike a pure attractive potential.
The two required conclusions must be physical (attractive/repulsive regions, short range/saturation) — merely describing the shape earns fewer marks.

Real-world link

The short range and saturation of this force explain why nuclear binding energy per nucleon is nearly constant (~8 MeV) across the periodic table — the backbone of both nuclear stability and energy release in fission and fusion.

Common Mistakes

1Drawing a purely attractive (Coulomb-like) curve with no repulsive core for $r < 0.8 \text{ fm}$ .
2Marking the equilibrium separation wrongly (e.g. at several fm) instead of $r_0 \approx 0.8 \text{ fm}$ where PE is minimum.
3Stating only the shape of the graph instead of two physical conclusions (attractive/repulsive nature and short range).

Interesting Facts

The ~100 MeV depth of this potential well is about ten million times deeper than the ~13.6 eV binding of the electron in hydrogen — a vivid measure of how mighty the strong nuclear force is.

The repulsive hard core at $r < 0.5 \text{ fm}$ is why nuclear matter is nearly incompressible; the same physics sets the density limit inside neutron stars.

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Frequently Asked Questions

What does the minimum of the potential-energy curve represent?

The minimum (about $-100 \text{ MeV}$ ) at $r_0 \approx 0.8 \text{ fm}$ marks the stable equilibrium separation of two nucleons, where the net force between them is zero. Nucleons naturally settle at this spacing inside a nucleus.

How does the graph show the nuclear force is short-ranged?

The potential energy rapidly approaches zero as the separation exceeds a few femtometres. Since force is the negative gradient of U, a flat curve means negligible force — so nucleons feel the strong force only over very short distances, with each nucleon interacting mainly with its nearest neighbours.