The mass density of a nucleus varies with mass number `A` as:

Options

A. `A^(2)`
B. `A`
C. constant
D. `1/A`

Answer

To solve the problem of how mass density varies with mass number \( A \), we can follow these steps:

### Step 1: Understand the formula for mass density
The mass density \( \rho \) of a nucleus is defined as the mass \( m \) divided by the volume \( V \):
\[ \rho = \frac{m}{V} \]

### Step 2: Express the volume of the nucleus
The volume \( V \) of a nucleus can be expressed in terms of its radius \( r \):
\[ V = \frac{4}{3} \pi r^3 \]

### Step 3: Relate the radius to the mass number \( A \)
The radius \( r \) of a nucleus is often approximated as:
\[ r = r_0 A^{1/3} \]
where \( r_0 \) is a constant.

### Step 4: Substitute the radius into the volume formula
Substituting the expression for \( r \) into the volume formula gives:
\[ V = \frac{4}{3} \pi (r_0 A^{1/3})^3 \]
This simplifies to:
\[ V = \frac{4}{3} \pi r_0^3 A \]

### Step 5: Substitute the volume back into the density formula
Now, substituting the volume back into the density formula:
\[ \rho = \frac{m}{\frac{4}{3} \pi r_0^3 A} \]

### Step 6: Express the mass in terms of \( A \)
For a nucleus, the mass \( m \) is approximately equal to the mass number \( A \):
\[ m = A \]
So, substituting this into the density equation gives:
\[ \rho = \frac{A}{\frac{4}{3} \pi r_0^3 A} \]

### Step 7: Simplify the expression
The \( A \) in the numerator and denominator cancels out:
\[ \rho = \frac{1}{\frac{4}{3} \pi r_0^3} \]

### Step 8: Conclusion
This shows that the mass density \( \rho \) is a constant value and does not depend on the mass number \( A \). Thus, we conclude that:
\[ \rho \text{ is constant and independent of } A. \]