Q.(a) Distinguish between isotopes and isobars, giving one example for each.

The key idea here is that nuclear radius scales with the cube root of the mass number. This comes from the fact that nuclei are roughly spherical and have nearly constant density — so volume (and thus radius cubed) is proportional to the number of nucleons.

1. Recall the nuclear radius formula For a nucleus, the radius is approximately

where $R_0$ is a constant (about $1.2 \times 10^{-15}$ m) and $A$ is the mass number. This relation holds because nuclear matter is incompressible: adding nucleons increases volume proportionally.

2. Set up the ratio If two nuclei have radii $R_1$ and $R_2$ with ratio

then using the formula:

3. Solve for the mass-number ratio Equate the two expressions:

Cube both sides:

4. Interpret the result …

Concept & Intuition

The key idea is that nuclear density is roughly constant for all atoms. Why? Because the volume of a nucleus is proportional to its mass number $A$ (the number of nucleons). Since mass is also proportional to $A$, the density $\rho = \frac{\text{mass}}{\text{volume}}$ becomes independent of $A$. This is a fundamental result of the liquid-drop model of the nucleus: nucleons pack together at a fixed density, like drops of an incompressible fluid.

Step-by-step reasoning

1. Recall the empirical formula for nuclear radius The radius $R$ of a nucleus is given by

where $R_0 \approx 1.2 \times 10^{-15}\,\text{m}$ is a constant, and $A$ is the mass number.

2. Express the volume in terms of $A$ The volume $V$ of a spherical nucleus is

So $V \propto A$.

3. Relate mass to mass number

   The mass of a nucleus is approximately $m \approx A \cdot u$, where $u$ is the atomic mass unit. Hence $m \propto A$.

4. Compute nuclear density

Therefore, nuclear density is the same for all nuclei.

5. Apply to the given ratios …

The nuclear radius follows $R=R_0A^{1/3}$ with $R_0\approx1.2$ fm. Solve for $A$: …

The key concept is the empirical nuclear radius formula: $R = R_0 A^{1/3}$, where $R_0$ is a constant (≈ 1.2 fm) and $A$ is the mass number. This arises because nuclear matter has roughly constant density — like a liquid drop — so volume is proportional to $A$, and radius is proportional to $A^{1/3}$.

Why this works:

If you double the number of nucleons, you double the volume, so the radius increases by a factor of $2^{1/3} \approx 1.26$. Here we know the radius for $A=27$ and want it for $A=125$. Instead of finding $R_0$, we can directly use the ratio.

1. Write the radius formula for both nuclei:

   For $^{27}\mathrm{Al}$: $R_{\mathrm{Al}} = R_0 (27)^{1/3}$

   For $^{125}\mathrm{Fe}$: $R_{\mathrm{Fe}} = R_0 (125)^{1/3}$

2. Take the ratio to eliminate $R_0$:

3. Simplify the cube roots: $125 = 5^3$ and $27 = 3^3$, so

4. Plug in the known radius: …

The key concept here is the nuclear radius formula. In nuclear physics, the radius of a nucleus is not proportional to its mass number $A$ directly, but rather to $A^{1/3}$. This is because the nucleus is roughly spherical and its volume is proportional to the number of nucleons (protons and neutrons). Since volume scales as $r^3$, the radius scales as $A^{1/3}$. The standard empirical formula is:

where $R_0$ is a constant (about $1.2 \times 10^{-15}$ m). For comparing two nuclei, the constant cancels out.

1. Write the radii for both elements. For element X with mass number $A_X = 232$:

For element Y with mass number $A_Y = 29$:

2. Take the ratio of the radii. The constant $R_0$ cancels, so:

3. Simplify the fraction inside the cube root.

(Since $29 \times 8 = 232$.)

4. Take the cube root. $$$$

Concept: nuclear radius R = R0 * A^(1/3), where A is the mass number (nuclear density is essentially constant). …

Nuclear radius: $R=R_0A^{1/3}$, so volume $V=\tfrac43\pi R^3=\tfrac43\pi R_0^3 A\propto A$.

Mass $\approx A m_N\propto A$. Therefore density: …