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 find the relation of mass density of a nucleus with mass number \( A \), we can follow these steps:

### Step 1: Understand the volume of the nucleus
Assuming the nucleus is spherical, the volume \( V \) of the nucleus can be expressed as:
\[ V = \frac{4}{3} \pi r^3 \]

### Step 2: Define the radius of the nucleus
The radius \( r \) of the nucleus is related to the mass number \( A \) by the formula:
\[ r = r_0 A^{1/3} \]
where \( r_0 \) is a constant (approximately \( 1.2 \, \text{fm} \)).

### Step 3: Substitute the radius into the volume formula
Substituting \( r \) 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 4: Calculate the mass of the nucleus
The mass \( m \) of the nucleus can be expressed as:
\[ m = A \cdot m_n \]
where \( m_n \) is the mass of a nucleon (approximately \( 1.67 \times 10^{-27} \, \text{kg} \)).

### Step 5: Determine the density of the nucleus
The density \( \rho \) of the nuclear matter is defined as the mass divided by the volume:
\[ \rho = \frac{\text{mass of nucleus}}{\text{volume of nucleus}} = \frac{m \cdot A}{V} \]
Substituting the expressions for mass and volume, we get:
\[ \rho = \frac{A \cdot m_n}{\frac{4}{3} \pi r_0^3 A} \]

### Step 6: Simplify the density expression
This simplifies to:
\[ \rho = \frac{3 m_n}{4 \pi r_0^3} \]
Notice that the density \( \rho \) is independent of the mass number \( A \).

### Final Result
The mass density of a nucleus varies with mass number \( A \) as:
\[ \rho \approx \frac{3 m_n}{4 \pi r_0^3} \]
This shows that the density of nuclear matter is approximately constant for different nuclei.