Question:

The mass density of a nucleus varies with mass number $ A $ as

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The mass density of a nucleus is nearly independent of the mass number $ A $, which means that it does not change as the size of the nucleus increases.

Updated On: May 3, 2025

$ A^0 $
$ A^2 $
$ \frac{1}{A} $
$ \ln A $

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The Correct Option is A

Approach Solution - 1

To determine how the mass density of a nucleus varies with the mass number $ A $, consider the definition of mass density in terms of mass and volume. The mass density $ \rho $ is given by:

$ \rho = \frac{m}{V} $

Where $ m $ is the mass of the nucleus and $ V $ is its volume. For most nuclei, the mass $ m $ can be approximated as proportional to the mass number $ A $, assuming that each nucleon (proton or neutron) has roughly the same mass. Mathematically, this can be expressed as:

$ m \propto A $

The volume $ V $ of a nucleus is related to its radius $ R $, which in turn is proportional to the cube root of the mass number $ A $, i.e.,

$ R \propto A^{1/3} $

The volume $ V $ of a sphere (which is a simplification for the shape of a nucleus) is given by:

$ V = \frac{4}{3} \pi R^3 $

Substituting the proportionality for $ R $, the volume $ V $ becomes:

$ V \propto (A^{1/3})^3 = A $

Therefore, the mass density $ \rho $ becomes:

$ \rho = \frac{m}{V} \propto \frac{A}{A} = A^0 $

This shows that the mass density of a nucleus is independent of the mass number $ A $ and is constant.

Hence, the correct option is: $ A^0 $

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Approach Solution -2

To determine how the mass density of a nucleus varies with the mass number $ A $, we must first understand the relationship between these quantities. The density $ \rho $ of a nucleus is given by:

$\rho = \frac{m}{V}$

where $ m $ is the mass of the nucleus and $ V $ is its volume. The mass of a nucleus is approximately proportional to the mass number $ A $, as the mass of the nucleus is primarily determined by its protons and neutrons.

The volume $ V $ of the nucleus is proportional to the cube of the radius $ R $ of the nucleus. The radius $ R $ of a nucleus is often approximated by the empirical formula:

$R = R_0 A^{1/3}$

where $ R_0 $ is a constant. This leads to the volume:

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

Thus, the volume is proportional to $ A $.

Substituting back into the density equation:

$\rho = \frac{m}{V} \approx \frac{A}{A} = 1$

This implies that the mass density of a nucleus is a constant value, independent of the mass number $ A $. Therefore, the mass density varies with mass number $ A $ as:

$A^0$

The correct answer is $ A^0 $, which signifies that the nuclear density is independent of the mass number.