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Q13.10:
From the relation \(R = R_{0}A^{\frac{1}{3}}\), where \(R_{0}\) is a constant and \(A\) is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

Solution :

Step 1: Establishing the Nuclear Radius and Volume

Given the empirical relationship for the radius of a nucleus as a function of the mass number $A$: $R = R_{0}A^{\frac{1}{3}}$ where $R_{0} \approx 1.2 \times 10^{-15} \, \text{m}$ (or $1.2 \, \text{fm}$).

Assuming the nucleus to be a sphere of volume $V$, we apply the geometric formula for the volume of a sphere: $V = \frac{4}{3}\pi R^3$ [By geometric definition of a sphere]

Substituting the expression for $R$: $V = \frac{4}{3}\pi (R_{0}A^{\frac{1}{3}})^3$ $V = \frac{4}{3}\pi R_{0}^3 A$ [Algebraic simplification using $(x^a)^b = x^{ab}$]

Step 2: Defining Nuclear Mass

The mass of a nucleus $M$ is approximately equal to the product of the mass number $A$ and the mass of a single nucleon $m$ (where $m \approx 1.67 \times 10^{-27} \, \text{kg}$): $M \approx A \cdot m$

Step 3: Calculating Nuclear Matter Density

Density $\rho$ is defined as mass per unit volume: $\rho = \frac{M}{V}$

Substituting the expressions derived in Step 1 and Step 2: $\rho = \frac{A \cdot m}{\frac{4}{3}\pi R_{0}^3 A}$

Observe that the term $A$ appears in both the numerator and the denominator. By canceling $A$: $\rho = \frac{m}{\frac{4}{3}\pi R_{0}^3}$ [Canceling common variable $A \neq 0$]

Step 4: Analytical Conclusion

Since $m$ (nucleon mass), $\pi$, and $R_{0}$ (constant) are all fundamental constants independent of the specific nucleus chosen, the resulting expression for $\rho$ contains no dependence on the mass number $A$.

R Constant Density ρ ≈ 2.3 × 10¹⁷ kg/m³ ρ = M / V = (Am) / (4/3 π R₀³ A) = constant

Data Summary Table

Variable Physical Significance
$A$ Mass Number (Total Nucleons)
$R$ Nuclear Radius $\propto A^{1/3}$
$V$ Nuclear Volume $\propto A$
$\rho$ Nuclear Density (Mass/Volume)

Final Solution: Because the nuclear volume is directly proportional to the mass number ($V \propto A$), the mass number cancels out when calculating density, proving that nuclear matter density $\rho = \frac{3m}{4\pi R_{0}^3}$ is independent of $A$ and is nearly constant for all nuclei.


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