The nucleus is approximately spherical in shape. Then the surface area of a nucleus having mass number A varie

I Multiple Choice Questions · Q10

A

(a)

A^{2/3}

(b)

A^{4/3}

(c)

A^{1/3}

(d)

A^{5/3}

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Concept understanding — Nuclear Density Calculation

Nuclear Density: Why All Nuclei Are Almost Equally Dense

Imagine you have a bag of marbles. If you pack them tightly, the density of the bag depends only on the marbles themselves — not on how many you put in. The nucleus behaves the same way. That's the core idea.

The Intuition

An atom's nucleus is made of protons and neutrons (collectively called nucleons). These nucleons are held together by the strong nuclear force, which is extremely short-ranged. Think of it like magnets: each nucleon only "feels" its immediate neighbours. So adding more nucleons doesn't compress the inner ones — it just adds a new layer on the outside.

This means the nucleus grows in volume proportionally to the number of nucleons. Double the number of nucleons, double the volume. And since mass also doubles, the density stays constant.

The Precise Statement

The nuclear radius $R$ is experimentally found to follow:

R = R_0 A^{1/3}

where:

$A$ = mass number (total protons + neutrons)
$R_0 \approx 1.2 \times 10^{-15} \text{ m}$ (a constant, about 1.2 femtometres)

R = R_0 A^{1/3}

This is the nuclear radius formula. It's not a guess — it comes from scattering experiments where high-energy electrons or alpha particles bounce off nuclei.

Deriving the Density

The nucleus is roughly spherical, so its volume is:

V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A

Notice: volume is directly proportional to $A$ . The mass of the nucleus is approximately $m \approx A \times (1.67 \times 10^{-27} \text{ kg})$ (mass of one nucleon). So density:

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

The $A$ cancels out completely. The density is a constant — independent of the nucleus size.

Important

Nuclear density is independent of mass number $A$ . All nuclei have approximately the same density.

The Numerical Value

Plug in the numbers:

$m_{\text{nucleon}} \approx 1.67 \times 10^{-27} \text{ kg}$
$R_0 \approx 1.2 \times 10^{-15} \text{ m}$

\rho = \frac{1.67 \times 10^{-27}}{\frac{4}{3}\pi (1.2 \times 10^{-15})^3} \approx 2.3 \times 10^{17} \text{ kg/m}^3

That's about 230 million tonnes per cubic centimetre. To put it in perspective: a sugar-cube-sized piece of nuclear matter would weigh as much as 230 million cars.

Note

| Object | Density (kg/m³) |

|--------|-----------------|

| Water | $10^3$ |

| Earth (average) | $5.5 \times 10^3$ |

| White dwarf star | $10^9$ |

| Atomic nucleus | $2.3 \times 10^{17}$ |

Why This Matters

This constancy of density tells us something profound: the strong nuclear force saturates. Each nucleon only interacts with its nearest neighbours, not with the whole nucleus. If the force were long-range (like gravity), density would increase with size. It doesn't — so the force is short-range. …

Why this formula?

Why Nuclear Density is Constant — The Reasoning

The most striking result about nuclear density is that it is roughly the same for all nuclei, regardless of size. This is not obvious — why wouldn't a larger nucleus be denser? The answer lies in how nuclear force works and how nucleons pack together.

Step 1: The nuclear volume formula

Experiments show that the radius of a nucleus is given by:

R = R_0 A^{1/3}

where $R_0 \approx 1.2 \times 10^{-15}\ \text{m}$ (1.2 fm) and $A$ is the mass number (total number of protons + neutrons).

Note

The $A^{1/3}$ dependence is the key. It means volume grows linearly with $A$ , not faster.

Why $A^{1/3}$ ? Because nucleons are packed as tightly as possible — like spheres in a close-packed arrangement. If you double the number of nucleons, you need to double the volume, so the radius must increase by $2^{1/3}$ .

Step 2: Volume from the radius

Assuming the nucleus is a sphere:

V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A

Notice: the $A^{1/3}$ cube gives $A$ directly. So volume is proportional to $A$ .

Step 3: Mass of the nucleus

The mass of the nucleus is approximately:

M \approx A \cdot m_{\text{nucleon}}

where $m_{\text{nucleon}} \approx 1.67 \times 10^{-27}\ \text{kg}$ (the average mass of a proton or neutron). The small mass defect from binding energy is negligible for this calculation.

Step 4: Density

Nuclear density $\rho$ is mass divided by volume:

\rho = \frac{M}{V} = \frac{A \cdot m_{\text{nucleon}}}{\frac{4}{3}\pi R_0^3 A} = \frac{m_{\text{nucleon}}}{\frac{4}{3}\pi R_0^3}

Important

The $A$ cancels completely. Nuclear density is independent of the nucleus size.

Step 5: The numerical value

Plugging in the numbers:

\rho = \frac{1.67 \times 10^{-27}}{\frac{4}{3}\pi (1.2 \times 10^{-15})^3} \approx 2.3 \times 10^{17}\ \text{kg/m}^3

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

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Step 1. The nucleus is treated as a sphere of radius $R=R_0A^{1/3}$ .

Step 2. The surface area of a sphere is $S=4\pi R^2$ , so

$S=4\pi(R_0A^{1/3})^2=4\pi R_0^2A^{2/3}\propto A^{2/3}$ …

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Q.The nucleus is approximately spherical in shape. Then the surface area of a nucleus having mass number AAA varies as

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Q.The nucleus is approximately spherical in shape. Then the surface area of a nucleus having mass number $A$ varies as

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