Q.Given the mass of iron nucleus is 55.85 a.m.u. and A = 56, find the nuclear density. (Given : 1 a.m.u. = 1.67×10⁻²⁷ kg, R₀ = 1.2×10⁻¹⁵ m.)

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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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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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 …

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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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 …

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).

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

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

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 $\lambda_p/\lambda_e=m_e/m_p\approx1/1836$.

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 …