JEE Mains Physics · Nuclei

Nuclear Size, Mass Defect and Binding Energy

A nucleus has radius R₀A^(1/3), so every nucleus has the same density; its mass is less than the mass of its free nucleons, and that missing mass times c² is the binding energy that holds it together.

All Nuclei notes JEE Mains Physics strategy

Why this matters

Thirty PYQs, six of them numerical, and seven from 2026, more than any other page in this chapter. Fourteen use the radius rule: a ratio of radii or volumes, a density that never changes, the speeds of two fragments. Eight find a binding energy from masses, and eight test the binding-energy curve and the nuclear force in words.

Concept 1 of 3: Nuclear radius and constant density

Nucleons pack like marbles in a bag: each takes the same small volume. So the volume of a nucleus grows in step with its mass number A, and the radius grows only as the cube root of A. Mass and volume both grow as A, so their ratio, the density, is the same for every nucleus.

Definition

$R = R_0A^{1/3}$ , with $R_0 \approx 1.2$ fm (1 fm = $10^{-15}$ m).
Volume $\propto A$ , surface area $\propto A^{2/3}$ , radius $\propto A^{1/3}$ . In ratios: $\left(\dfrac{R_1}{R_2}\right)^{3} = \dfrac{A_1}{A_2}$ .
Density $\rho = \dfrac{Am}{\tfrac{4}{3}\pi R_0^{3}A} = \dfrac{3m}{4\pi R_0^{3}} \approx 2.3 \times 10^{17}\ \text{kg/m}^{3}$ . A cancels, so every nucleus has the same density.
Only protons and neutrons count in A. Electrons add nothing to the mass number.
A nucleus at rest splitting in two: momentum stays zero, so $A_1v_1 = A_2v_2$ . Then $\dfrac{v_1}{v_2} = \dfrac{A_2}{A_1}$ and $\dfrac{R_1}{R_2} = \left(\dfrac{A_1}{A_2}\right)^{1/3}$ .

Nuclear radius

R = R_0A^{1/3}, \qquad \rho = \frac{3m}{4\pi R_0^{3}}\ \text{(same for all A)}

$R_0$ a constant, about 1.2 fm
$A$ mass number (protons + neutrons)
$m$ mass of one nucleon

Worked example

The nucleus

^{27}\text{Al}

has radius 3.6 fm. Find

R_0

, the radius of a nucleus with

A = 125

, and the ratio of their surface areas.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q105Moderate

Example 1 · Nuclei · Nuclear Size, Mass Defect and Binding Energy

The mass number of nucleus having radius equal to half of the radius of nucleus with mass number 192 is:

Any statement that orders nuclear densities is false

Bismuth is not denser than lithium. The A in the mass cancels the A in the volume, so a statement ranking nuclei by density, or saying density grows with A, is wrong. The reason

R \propto A^{1/3}

is true; the claim built on it is not.

Cube the radius ratio, do not cube-root it twice

Halving the radius divides A by 8, not by 2. Go from radii to mass numbers by cubing, and from mass numbers to radii by taking the cube root.

Absorbed electrons do not change A

If a nucleus captures protons, neutrons and electrons, only the protons and neutrons raise A. Count the nucleons, then apply

A^{1/3}

A^{2/3}

Drill 13 more on nuclear radius and constant density

Concept 2 of 3: Mass defect and binding energy

Weigh the free protons and neutrons, then weigh the nucleus they make: the nucleus is lighter. The missing mass left as energy when the nucleus formed. Putting that energy back, Δm c², is what it takes to pull the nucleus apart. That is the binding energy.

Definition

Mass defect $\Delta m = Zm_p + (A - Z)m_n - M_{\text{nucleus}}$ .
Binding energy $BE = \Delta m\,c^{2}$ ; with masses in u, $1\ \text{u} = 931.5\ \text{MeV}/c^{2}$ .
With atomic masses, use the mass of a hydrogen atom, $m(^{1}\text{H}) = 1.007825$ u, in place of $m_p$ . The Z electrons then cancel.
BE per nucleon, $BE/A$ , measures how tightly bound a nucleus is. Compare stability with it, not with total BE.
The other direction: $\Delta m = BE/c^{2}$ . Energy in joules divided by $9 \times 10^{16}$ gives kilograms.
Neutron separation energy: $S_n = [M(A - 1) + m_n - M(A)]c^{2}$ , the energy to pull out one neutron.

Binding energy

BE = \left[Zm_p + (A - Z)m_n - M\right]c^{2}, \qquad 1\ \text{u}\,c^{2} = 931.5\ \text{MeV}

Worked example

Find the binding energy and the binding energy per nucleon of

^{7}_{3}\text{Li}

. Atomic mass of

^{7}\text{Li}

= 7.01600 u;

m(^{1}\text{H})

= 1.00783 u;

m_n

= 1.00867 u; 1 u = 931.5 MeV/

c^{2}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q19Moderate

Example 2 · Nuclei · Nuclear Size, Mass Defect and Binding Energy

The binding energy per nucleon of

\ _{83}^{209}Bi

\_\_\_\_

MeV .[Take

m\left( \ _{83}^{209}Bi \right)= 208.980388u,m_{p}= 1.007825u

m_{n}= 1.008665u,1u = 931MeV/c^{2}

]

Atom mass or nucleus mass

Tables give atomic masses, which include the electrons. Pair an atomic mass with

m(^{1}\text{H})

for the protons so the electrons cancel. Mixing an atomic mass with bare proton masses leaves Z electron masses in the defect.

Bound mass is the smaller one

The defect is (free nucleons) − (nucleus). Written the other way round, the binding energy comes out negative.

Per nucleon or in total

Read whether the question wants BE or BE/A. Stability is compared with BE per nucleon; the energy to break the whole nucleus is the total.

Drill 7 more on mass defect and binding energy

Concept 3 of 3: The binding-energy curve and the nuclear force

Plot BE per nucleon against A. It climbs fast for light nuclei, peaks near iron at A ≈ 56, and then falls slowly. In the long middle it is almost flat, because the nuclear force reaches only the nearest neighbours: adding nucleons adds the same bond energy each time. Any change that moves nuclei toward the peak releases energy.

Definition

Nuclear force: the strongest force, short ranged (a few fm), attractive at those distances and repulsive when nucleons get too close. It is charge independent (p–p, n–n and n–p alike), spin dependent, not inverse-square, and it saturates.
Liquid-drop terms in the binding energy: volume term $\propto A$ (adds), surface term $\propto A^{2/3}$ (subtracts, since surface nucleons have fewer neighbours), Coulomb term $\propto Z(Z - 1)/A^{1/3}$ (subtracts).
Isotopes: same Z, different A ( $^{12}\text{C}$ , $^{14}\text{C}$ ). Isobars: same A, different Z ( $^{40}_{18}\text{Ar}$ , $^{40}_{20}\text{Ca}$ ). Isotones: same number of neutrons ( $^{13}_{6}\text{C}$ , $^{14}_{7}\text{N}$ ).
Nuclei with lower BE per nucleon tend to change into nuclei with higher BE per nucleon: heavy ones by fission, light ones by fusion.

Part of the curve	Mass number	BE per nucleon	What it means
Lightest nuclei	Below about 20	Low and uneven: about 1.1 MeV for $^{2}\text{H}$ , a spike near 7 MeV for $^{4}\text{He}$	Fusing two light nuclei raises BE per nucleon and releases energy.
Flat middle	About 30 to 170	Nearly constant, about 8 MeV	The force is short ranged and saturates: each nucleon binds only to its neighbours. Flat because the force is SHORT ranged. A reason that says long range is false.
Peak	Near 56 (iron)	Highest, about 8.8 MeV	The most tightly bound nuclei; neither fission nor fusion releases energy from them.
Heavy nuclei	Above about 170	Falls slowly, to about 7.6 MeV for uranium	Coulomb repulsion grows; splitting into two middle nuclei releases energy.

Energy is released whenever the products sit higher on this curve than what you started with.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q91Moderate

Example 3 · Nuclei · Nuclear Size, Mass Defect and Binding Energy

Given below are two statements: one is labelled as Assertion

A

and the other is labelled as Reason

R

Assertion A: The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range 30 to 170. Reason R: Nuclear force is short ranged. In the light of the above statements, choose the correct answer from the options given below Option:

Heavier is not always more tightly bound

BE per nucleon rises only up to iron. Beyond A ≈ 56 it falls. A statement that says it grows with mass for all nuclei is false.

Isobars share A, isotopes share Z

Iso-BAR: same mass number (the bar on a balance weighs mass). Iso-TOPE: same place in the periodic table, so same Z. Isotones share N.

Stability is judged per nucleon

A bigger nucleus has a bigger total binding energy almost always. Compare BE per nucleon to say which nucleus is more stable.

Drill 7 more on the binding-energy curve and the nuclear force

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Nuclear radius and constant density
Nuclear radius
$R = R_0A^{1/3}, \qquad \rho = \frac{3m}{4\pi R_0^{3}}\ \text{(same for all A)}$
Mass defect and binding energy
Binding energy
$BE = \left[Zm_p + (A - Z)m_n - M\right]c^{2}, \qquad 1\ \text{u}\,c^{2} = 931.5\ \text{MeV}$

Reference tables (1)

The binding-energy curve and the nuclear force4 rows

Part of the curve	Mass number	BE per nucleon	What it means
Lightest nuclei	Below about 20	Low and uneven: about 1.1 MeV for $^{2}\text{H}$ , a spike near 7 MeV for $^{4}\text{He}$	Fusing two light nuclei raises BE per nucleon and releases energy.
Flat middle	About 30 to 170	Nearly constant, about 8 MeV	The force is short ranged and saturates: each nucleon binds only to its neighbours. Flat because the force is SHORT ranged. A reason that says long range is false.
Peak	Near 56 (iron)	Highest, about 8.8 MeV	The most tightly bound nuclei; neither fission nor fusion releases energy from them.
Heavy nuclei	Above about 170	Falls slowly, to about 7.6 MeV for uranium	Coulomb repulsion grows; splitting into two middle nuclei releases energy.

Energy is released whenever the products sit higher on this curve than what you started with.

Watch out for (9)

Any statement that orders nuclear densities is false
→ Nuclear radius and constant density
Cube the radius ratio, do not cube-root it twice
→ Nuclear radius and constant density
Absorbed electrons do not change A
→ Nuclear radius and constant density
Atom mass or nucleus mass
→ Mass defect and binding energy
Bound mass is the smaller one
→ Mass defect and binding energy
Per nucleon or in total
→ Mass defect and binding energy
Heavier is not always more tightly bound
→ The binding-energy curve and the nuclear force
Isobars share A, isotopes share Z
→ The binding-energy curve and the nuclear force
Stability is judged per nucleon
→ The binding-energy curve and the nuclear force

Test yourself on Nuclei

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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