"For a nucleus of mass number A and radius R, the mass density of nucleus can be represented as"

Question

Accepted Answer

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The radius $ R $ of a nucleus is given by the empirical relation:

$R = r_0 A^{\frac{1}{3}},$

where $ r_0 $ is a constant and $ A $ is the mass number.

The volume $ V $ of the nucleus is approximately that of a sphere:

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

Substituting the expression for $ R $ into the volume formula:

$V = \frac{4}{3}\pi \left(r_0 A^{\frac{1}{3}}\right)^3 = \frac{4}{3}\pi r_0^3 A.$

The mass density $ \rho $ is defined as the mass (which is proportional to $ A $) divided by the volume:

$$ \rho = \frac{A}{V} = \frac{A}{\frac{4}{3}\pi r_0^3 A} = \frac{1}{\frac{4}{3}\pi r_0^3}. $$

Notice that $ A $ cancels out in the numerator and denominator, so the density is:

$\rho = \text{constant},$

meaning it is independent of the mass number $ A $.

Thus, the correct option is:

Option B: Independent of A.

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For a nucleus of mass number A and radius R, the mass density of nucleus can be represented as