To solve this question, we need to understand how the mass density of a nucleus relates to its mass number \( A \) and its radius \( R \).
The mass density \( \rho \) of a nucleus is given by the formula:
\(\rho = \frac{\text{mass of nucleus}}{\text{volume of nucleus}}\)
We know that the mass of a nucleus is approximately proportional to its mass number \( A \), because the mass number represents the sum of protons and neutrons, which have nearly equal mass.
The volume \( V \) of a spherical nucleus is calculated using the formula:
\(V = \frac{4}{3} \pi R^3\)
For a nucleus, it is empirically found that the radius \( R \) is related to the mass number \( A \) by the equation:
\(R = R_0 A^{1/3}\)
where \( R_0 \) is a constant. Thus, the volume \( V \) can be expressed as:
\(V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A\)
Substituting this into the formula for density, we get:
\(\rho = \frac{A}{\frac{4}{3} \pi R_0^3 A}\)
Upon simplification, it becomes:
\(\rho = \frac{1}{\frac{4}{3} \pi R_0^3}\)
From this expression, we can see that the density \( \rho \) is independent of \( A \) since the \( A \) terms cancel each other out.
Therefore, the correct answer is:
Independent of \( A \)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

In a nuclear fission process, a high mass nuclide (A ≈ 236) with binding energy 7.6 MeV/Nucleon dissociated into middle mass nuclides (A ≈ 118), having binding energy of 8.6 MeV/Nucleon. The energy released in the process would be ____ MeV.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)