Size of the Nucleus & Nuclear Density

Measuring the Unmeasurable

How do you measure something $10^{-14}$ m across? You throw things at it.

From closest approach to an upper bound

Geiger and Marsden's 5.5 MeV alphas approached the gold nucleus to about $4.0 \times 10^{-14}$ m at closest — and since Rutherford's pure-Coulomb calculation matched the data perfectly, the positive charge must sit entirely within that distance: the gold nucleus is smaller than $4.0 \times 10^{-14}$ m.

Use higher-energy alphas and the closest approach shrinks — until, at some energy, the results deviate from Rutherford's Coulomb-only predictions: the alpha has begun to feel the short-range nuclear force. The distance where deviations set in reveals the nuclear size.

Electron scattering: the precision tool

The accurate sizes come from scattering fast electrons off nuclei. (Electrons feel only the electromagnetic force — no nuclear-force complications — so they cleanly map the charge distribution.)

The radius rule

All measurements fit one beautiful formula:

$$\boxed{R = R_0A^{1/3}}, \qquad R_0 = 1.2 \times 10^{-15} \text{ m} = 1.2 \text{ fm}$$

Radius grows as the cube root of the mass number: gold (A = 197) is only $\left(\dfrac{197}{107}\right)^{1/3} \approx 1.23$ times wider than silver (A = 107).

The Astonishing Consequence: Constant Density

The cube-root law hides a deep fact. Nuclear volume:

$$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3\,A \propto A$$

Volume is proportional to the number of nucleons. So the density,

$$\rho = \frac{\text{mass}}{\text{volume}} \approx \frac{A \times m_{nucleon}}{\frac{4}{3}\pi R_0^3 A} = \frac{3m_{nucleon}}{4\pi R_0^3}$$

is independent of A — the A cancels! Every nucleus, from helium to uranium, is a droplet of the same universal liquid:

$$\rho_{nuclear} \approx 2.3 \times 10^{17} \text{ kg/m}^3$$

Compare water's $10^3$ kg/m³ — nuclear matter is ~$10^{14}$ times denser. No contradiction: ordinary matter is mostly the empty space of atoms; the nucleus is matter with the emptiness squeezed out.

NCERT Example 13.1: iron's density

With $m_{Fe} = 55.85$ u $= 9.27 \times 10^{-26}$ kg and A = 56:

$$\rho = \frac{9.27 \times 10^{-26}}{\frac{4\pi}{3}(1.2 \times 10^{-15})^3 \times 56} = 2.29 \times 10^{17} \text{ kg/m}^3$$

Neutron stars have comparable density — matter compressed until the star resembles one giant nucleus.

[JEE Tip] Ratio drills: $\dfrac{R_1}{R_2} = \left(\dfrac{A_1}{A_2}\right)^{1/3}$ — NCERT Exercise 13.4 (gold vs silver): $\left(\dfrac{197}{107}\right)^{1/3} \approx 1.23$. And density questions answer themselves: constant, same for all nuclei. If a 'nuclear density depends on A' option appears, it's wrong.

[NEET Important] Points to Ponder nuance: electron scattering senses the charge distribution; alpha scattering senses nuclear matter — the two radii differ slightly. Asked as assertion-reason.

Solved Examples

Example 1: Nuclear density of iron (NCERT Example 13.1)

Given the iron nucleus mass 55.85 u and A = 56, find the nuclear density.

Solution:

1. Mass: $m = 55.85 \times 1.66 \times 10^{-27} = 9.27 \times 10^{-26}$ kg.
2. Volume: $V = \frac{4\pi}{3}R_0^3A = \frac{4\pi}{3}(1.2 \times 10^{-15})^3 \times 56$.
3. Density: $\rho = \dfrac{9.27 \times 10^{-26}}{\frac{4\pi}{3}(1.728 \times 10^{-45})(56)} = 2.29 \times 10^{17}$ kg/m³.
4. Takeaway: the same number emerges for ANY nucleus — density is A-independent by construction.

Example 2: Gold vs silver radii (NCERT Exercise 13.4)

Find the ratio of the nuclear radii of $^{197}$Au and $^{107}$Ag.

Solution:

1. Rule: $\dfrac{R_{Au}}{R_{Ag}} = \left(\dfrac{197}{107}\right)^{1/3}$.
2. Compute: $\left(1.84\right)^{1/3} \approx 1.23$.
3. Answer: gold's nucleus is only ~23% wider despite nearly double the nucleons — the cube root at work.

Example 3: Radius of a uranium nucleus [Board Numerical]

Estimate the radius of $^{238}$U.

Solution:

1. Formula: $R = R_0A^{1/3} = 1.2 \times (238)^{1/3}$ fm.
2. Cube root: $238^{1/3} \approx 6.2$.
3. Answer: R ≈ 7.4 fm = $7.4 \times 10^{-15}$ m.
4. Takeaway: even the heaviest natural nucleus is under 8 fm — all of nuclear physics happens within ~10 fm.

Example 4: Constant density, shown generally (NCERT Exercise 13.10)

From $R = R_0A^{1/3}$, show that nuclear matter density is independent of A.

Solution:

1. Mass: ≈ $A\,m$ where m ≈ 1 u (a nucleon's mass).
2. Volume: $\frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3A$.
3. Density: $\rho = \dfrac{Am}{\frac{4}{3}\pi R_0^3A} = \dfrac{3m}{4\pi R_0^3}$ — A cancels.
4. Numerically: $\rho = \dfrac{3 \times 1.66 \times 10^{-27}}{4\pi(1.2 \times 10^{-15})^3} \approx 2.3 \times 10^{17}$ kg/m³. ∎

Example 5: Which nucleus has double the radius? [JEE Numerical]

A nucleus has mass number A = 27 and radius R. Which mass number gives radius 2R?

Solution:

1. Rule: $R \propto A^{1/3}$, so doubling R needs A to grow by $2^3 = 8$.
2. Compute: $A' = 8 \times 27 = 216$.
3. Takeaway: radii double only when nucleon numbers octuple — cube the radius factor.

Example 6: Volume ratio [NEET Numerical]

Find the ratio of the volumes of $^{64}$Zn and $^{8}$Be nuclei… using only mass numbers.

Solution:

1. Volume ∝ A: $\dfrac{V_{Zn}}{V_{Be}} = \dfrac{64}{8} = 8$.
2. Radius check: $\dfrac{R_{Zn}}{R_{Be}} = \left(\dfrac{64}{8}\right)^{1/3} = 2$; volume ratio $2^3 = 8$. ✔
3. Takeaway: for volumes skip the cube root entirely — V ∝ A directly.

Example 7: Mass of a teaspoon of nuclear matter [Conceptual Numerical]

Estimate the mass of 5 mL (a teaspoon) of nuclear-density matter.

Solution:

1. Volume: 5 mL = $5 \times 10^{-6}$ m³.
2. Mass: $m = \rho V = 2.3 \times 10^{17} \times 5 \times 10^{-6}$.
3. Answer: ≈ $10^{12}$ kg — a billion tonnes in a teaspoon.
4. Takeaway: this is neutron-star matter; the number makes 'nuclear density is enormous' vivid and is a favourite talking-point question.

Example 8: When does Rutherford's formula fail? [Board Conceptual]

Why do very high-energy alpha particles deviate from Rutherford's scattering predictions, and what is learned from the deviation?

Solution:

1. Rutherford's analysis assumes pure Coulomb repulsion between alpha and nucleus.
2. Higher energy → smaller closest approach; once within a few fm, the short-range nuclear force acts, altering the scattering.
3. The distance at which deviations set in marks where the nuclear force begins — i.e. the nuclear size.
4. Takeaway: agreement with Coulomb-only = 'still outside the nucleus'; deviation = 'touched it'. The failure of the formula is itself the measurement.

Example 9: Radius from femtometre data [NEET Numerical]

The radius of $^{27}$Al is 3.6 fm. Predict the radius of $^{125}$Te.

Solution:

1. Ratio: $\dfrac{R_{Te}}{R_{Al}} = \left(\dfrac{125}{27}\right)^{1/3} = \dfrac{5}{3}$.
2. Compute: $R_{Te} = 3.6 \times \dfrac{5}{3} = 6.0$ fm.
3. Takeaway: exam setters choose perfect cubes (27, 125) — spot them and the cube root is mental arithmetic.

Example 10: Two probes, two radii [JEE Conceptual]

Why do electron scattering and alpha scattering give slightly different nuclear radii?

Solution:

1. Electrons interact electromagnetically only → they map the charge (proton) distribution.
2. Alphas feel the nuclear force too → they sense the nuclear matter distribution (protons + neutrons).
3. The two distributions differ slightly, so the extracted radii differ — NCERT Points to Ponder 2.
4. Takeaway: 'what force the probe feels decides what the probe measures' — a neat assertion-reason discriminator.