(a) Consider the following nuclides: $^{12}_{6}\text{C},\ ^{198}_{80}\text{Hg},\ ^{14}_{6}\text{C},\ ^{197}_{79}\text{Au}$. Group them into isotopes and isotones. (b) How does the size of a nucleus depend on its mass number $A$? Hence prove that the density of a nucleus is a constant, independent of $A$, for all nuclei.

> [!TLDR] > Isotopes share the same proton number $Z$; isotones share the same neutron number $N = A - Z$. For the given nuclides, $^{12}_{6}\text{C}$ and $^{14}_{6}\text{C}$ are isotopes; $^{198}_{80}\text{Hg}$ and $^{197}_{79}\text{Au}$ are isotones. Nuclear size scales as $R = R_0 A^{1/3}$, making density independent of $A$. --- ### Part (a) — Grouping into isotopes and isotones The key distinction is simple: - **Isotopes** have the same number of protons ($Z$) but different mass numbers ($A$). - **Isotones** have the same number of neutrons ($N = A - Z$) but different $Z$. Let’s list each nuclide with its proton number $Z$ and neutron number $N$: | Nuclide | $Z$ | $N = A - Z$ | |---------|-----|-------------| | $^{12}_{6}\text{C}$ | 6 | 6 | | $^{14}_{6}\text{C}$ | 6 | 8 | | $^{198}_{80}\text{Hg}$ | 80 | 118 | | $^{197}_{79}\text{Au}$ | 79 | 118 | 1. **Isotopes**: Look for same $Z$. Carbon-12 and carbon-14 both have $Z = 6$. So $^{12}_{6}\text{C}$ and $^{14}_{6}\text{C}$ are isotopes of carbon. 2. **Isotones**: Look for same $N$. Mercury-198 has $N = 118$; gold-197 also has $N = 118$. So $^{198}_{80}\text{Hg}$ and $^{197}_{79}\text{Au}$ are isotones. > [!WARNING] > A common mistake is to confuse isotones with isotopes. Isotones share neutron number, not proton number. Here, Hg and Au have different $Z$ but the same $N$ — that’s the isotone condition. --- ### Part (b) — Nuclear size and constant density The size of a nucleus is determined by the strong nuclear force, which has a very short range. Experiments (like Rutherford scattering) show that the nuclear radius $R$ follows: > [!FORMULA] > $$ R = R_0 A^{1/3} $$ > where $R_0 \approx 1.2 \times 10^{-15}\ \text{m}$ (1.2 fm) is a constant for all nuclei. This tells us that volume scales linearly with $A$: 1. **Volume of a nucleus** (treating it as a sphere): $$ V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A $$ 2. **Mass of a nucleus** is approximately $m \ap

(a) Consider the following nuclides: ¹²₆C,¹⁹⁸₈₀Hg,¹⁴₆C,¹⁹⁷₇₉Au. Group them into isotopes and isotones. (b) How

Question

^{12}_{6}\text{C},\ ^{198}_{80}\text{Hg},\ ^{14}_{6}\text{C},\ ^{197}_{79}\text{Au}

(b) How does the size of a nucleus depend on its mass number

A

? Hence prove that the density of a nucleus is a constant, independent of

A

, for all nuclei.

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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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Isotopes share the same proton number $Z$ ; isotones share the same neutron number $N = A - Z$ . For the given nuclides, $^{12}_{6}\text{C}$ and $^{14}_{6}\text{C}$ are isotopes; $^{198}_{80}\text{Hg}$ and $^{197}_{79}\text{Au}$ are isotones. Nuclear size scales as $R = R_0 A^{1/3}$ , making density independent of $A$ .

Part (a) — Grouping into isotopes and isotones

The key distinction is simple:

Isotopes have the same number of protons ( $Z$ ) but different mass numbers ( $A$ ).
Isotones have the same number of neutrons ( $N = A - Z$ ) but different $Z$ .

Let’s list each nuclide with its proton number $Z$ and neutron number $N$ :

Nuclide	$Z$	$N = A - Z$
$^{12}_{6}\text{C}$	6	6
$^{14}_{6}\text{C}$	6	8
$^{198}_{80}\text{Hg}$	80	118
$^{197}_{79}\text{Au}$	79	118

Isotopes: Look for same $Z$ . Carbon-12 and carbon-14 both have $Z = 6$ . So $^{12}_{6}\text{C}$ and $^{14}_{6}\text{C}$ are isotopes of carbon.
Isotones: Look for same $N$ . Mercury-198 has $N = 118$ ; gold-197 also has $N = 118$ . So $^{198}_{80}\text{Hg}$ and $^{197}_{79}\text{Au}$ are isotones.

Watch out

A common mistake is to confuse isotones with isotopes. Isotones share neutron number, not proton number. Here, Hg and Au have different $Z$ but the same $N$ — that’s the isotone condition.

Part (b) — Nuclear size and constant density

The size of a nucleus is determined by the strong nuclear force, which has a very short range. Experiments (like Rutherford scattering) show that the nuclear radius $R$ follows:

R = R_0 A^{1/3}

This tells us that volume scales linearly with $A$ :

Volume of a nucleus (treating it as 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

Mass of a nucleus is approximately $m \approx A \cdot m_p$ , where $m_p$ is the proton mass (neutrons have nearly the same mass). So $m \propto A$ .
Density $\rho$ is mass per volume:

$\rho = \frac{m}{V} \approx \frac{A m_p}{\frac{4}{3}\pi R_0^3 A} = \frac{m_p}{\frac{4}{3}\pi R_0^3}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set A1 markMCQ
$Q. The nuclear density is approximately (A) independent of mass number (B) directly proportional to mass number (C) inversely proportional to mass number (D) directly proportional to A^(1/3)$
›Reveal solutionSolution

Nuclear radius R = R₀A^(1/3), so volume ∝ A and mass ∝ A, making density independent of A.

The nuclear radius follows the empirical relation $R = R_0 A^{1/3}$ , where $R_0 \approx 1.2$ fm and $A$ is the mass number.

Volume of the nucleus: $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 A$ , so $V \propto A$ .

Mass of the nucleus: $M \approx A \times m_{nucleon}$ , so $M \propto A$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The nuclei have their mass numbers in the ratio of 1:3. The ratio of their nuclear densities would be$
(a) (3)^(1/3) : 1
(b) 1:1
(c) 1:3
(d) 3:1
›Reveal solutionSolution

Because the nuclear radius scales as R = R0 A^(1/3), the nuclear volume scales exactly as A, so density = mass/volume comes out essentially the same constant for every nucleus, regardless of A.

Nuclear radius: R = R0 A^(1/3), where R0 approx 1.2 fm is a constant.

Nuclear volume: V = (4/3) pi R^3 = (4/3) pi R0^3 A

Since nuclear mass is approximately proportional to A (mass number, roughly A times the nucleon mass), density:

rho = mass / volume is proportional to A / A = constant

…

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CBSE 2025Set ANNUAL1 mark
$Q. The density of nuclear matter is independent of the size of the nucleus. (T/F)$
›Reveal solutionSolution

This statement is True — nuclear density is essentially the same for all nuclei, independent of the mass number (size) of the nucleus.

The radius of a nucleus with mass number $A$ is given empirically by $R = R_0 A^{1/3}$ , where $R_0 \approx 1.2$ fm is a constant. The nuclear volume is then $V \propto R^3 \propto A$ , and since the nuclear mass is also proportional to $A$ (each nucleon has roughly the same mass), the density …

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CBSE 2024Set ANNUAL1 mark
$1:4$
›Reveal solutionSolution

Nuclear density is (almost exactly) the same for all nuclei, regardless of mass number.

Nuclear density is given by $\rho = \dfrac{\text{mass}}{\text{volume}} = \dfrac{mA}{\frac43\pi R^3}$ , where the nuclear radius scales as $R = R_0 A^{1/3}$ . Substituting:

$\rho = \frac{mA}{\frac43\pi (R_0A^{1/3})^3} = \frac{mA}{\frac43\pi R_0^3 A} = \frac{m}{\frac43\pi R_0^3}$

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Two nuclei have mass numbers in the ratio 1:3, the ratio of the nuclear densities are$
(a) 3:1
(b) 1:1
(c) 1:9
(d) 1:3
›Reveal solutionSolution

Nuclear density $\rho = \text{mass}/\text{volume} \propto A/A^3 \cdot 3 =$ constant, since the nuclear radius $R\propto A^{1/3}$ makes volume $\propto A$ — so it does not depend on mass number at all.

The empirical nuclear radius formula is $R = R_0 A^{1/3}$ , where $R_0\approx 1.2\,\text{fm}$ is a constant and $A$ is the mass number. The nuclear volume is

$V = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi R_0^3 A$

which is directly proportional to $A$ . The nuclear mass is also (to good approximation) proportional to $A$ , since $M \approx A\,u$ (each nucleon contributes roughly one atomic mass unit).

So the nuclear density is …

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CBSE 2023Set 55/1/11 markMCQ
$A$
(a) proportional to $A^{1/3}$
(b) proportional to $A^{2/3}$
(c) proportional to $A^{3}$
(d) independent of $A$
›Reveal solutionSolution
The mass density of a nucleus is roughly constant for all nuclei because both the mass and the volume scale with the mass number $A$ , making the density independent of $A$ . The correct option is (d).

The key idea here is that a nucleus behaves like a tiny, incompressible drop of nuclear matter. Its volume is proportional to the number of nucleons (protons and neutrons), and its mass is also proportional to that number. When you take the ratio, the $A$ cancels out.

Let’s see why this is true step by step.

What is mass number $A$ ?

$A$ is the total number of nucleons in the nucleus. The mass of a single nucleon is roughly $m_n \approx 1.67 \times 10^{-27}$ kg. So the mass of the nucleus is approximately $M \approx A \cdot m_n$ . This is a direct proportionality: $M \propto A$ .

How does the size of a nucleus scale with $A$ ?

Experiments (like Rutherford scattering) show that the radius $R$ of a nucleus follows the empirical formula:

$R = R_0 A^{1/3}$
where $R_0 \approx 1.2 \times 10^{-15}$ m (about 1.2 femtometers). This is a well-established result — the nuclear volume grows with the number of nucleons.

What is the volume of the nucleus? Treating the nucleus as a sphere:

$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$
So $V \propto A$ . The volume is directly proportional to the number of nucleons.

Now compute the density: …
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CBSE 2022Set ANNUAL1 mark
$Q. Two nuclei have mass numbers in the ratio 1 : 27. What is the ratio of nuclear density?$
›Reveal solutionSolution

Nuclear density is the same for all nuclei, so the ratio is 1 : 1.

The nuclear radius is $R = R_0 A^{1/3}$ , so the nuclear volume is

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

The nuclear mass is proportional to the mass number, $m \approx A\,m_{nucleon}$ . Hence the density is

$\rho = \frac{m}{V} = \frac{A\,m_{nucleon}}{\frac{4}{3}\pi R_0^3 A} = \frac{m_{nucleon}}{\frac{4}{3}\pi R_0^3},$ …

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CBSE 2020Set 55/1/11 mark
$^{27}_{13}\text{Al}$
(OR)
$A proton and an electron have equal speeds. Find the ratio of de Broglie wavelengths associated with them.$
›Reveal solutionSolution

Part (a): Since $R\propto A^{1/3}$ , $R_{\text{Cu}}=R_{\text{Al}}(64/27)^{1/3}=3.6\times\tfrac43=4.8\ \text{fermi}$ .

Part (b): With equal speeds $\lambda\propto1/m$ , so $\lambda_p/\lambda_e=m_e/m_p\approx1/1836$ .

Part (a)

The nucleus behaves like a constant-density drop, so its volume $\propto A$ and its radius follows
$R=R_0A^{1/3},\qquad R_0\approx1.2\ \text{fm}.$
Taking a ratio removes $R_0$ :
$\frac{R_{\text{Cu}}}{R_{\text{Al}}}=\left(\frac{A_{\text{Cu}}}{A_{\text{Al}}}\right)^{1/3}=\left(\frac{64}{27}\right)^{1/3}=\frac{4}{3}.$
With $R_{\text{Al}}=3.6\ \text{fm}$ :

$R_{\text{Cu}}=3.6\times\frac{4}{3}=4.8\ \text{fm}.$ …
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CBSE 2020Set ANNUAL1 markMCQ
$Q. The nucleus is approximately spherical in shape. Then the surface area of nucleus having mass number A varies as :$
(a) $A^{5/3}$
(b) $A^{2/3}$
(c) $A^{4/3}$
(d) $A^{1/3}$
›Reveal solutionSolution

Since the nuclear radius scales as $A^{1/3}$ , the surface area of the (spherical) nucleus scales as $(A^{1/3})^2=A^{2/3}$ .

Working

The empirical nuclear radius formula is

$R=R_0A^{1/3},\qquad R_0\approx1.2\ \text{fm}$

Treating the nucleus as a sphere, its surface area is …

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Q.(a) Consider the following nuclides: 612C, 80198Hg, 614C, 79197Au^{12}_{6}\text{C},\ ^{198}_{80}\text{Hg},\ ^{14}_{6}\text{C},\ ^{197}_{79}\text{Au}612​C, 80198​Hg, 614​C, 79197​Au. Group them into isotopes and isotones.

Part (a) — Grouping into isotopes and isotones

Part (b) — Nuclear size and constant density

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Q.(a) Consider the following nuclides: $^{12}_{6}\text{C},\ ^{198}_{80}\text{Hg},\ ^{14}_{6}\text{C},\ ^{197}_{79}\text{Au}$ . Group them into isotopes and isotones.

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