Assuming the radius of a nucleus to be equal to `R=1.3 A^(1//3)xx10^(-15)m`. Where `A` is its mass number, evaluate the density of nuclei and the number of nucleons per unit volume of the nucleus. Take mass of one nucleon `=1.67xx10^(-27)kg`

Explanation

To solve the problem step by step, we will evaluate the density of nuclei and the number of nucleons per unit volume of the nucleus.

### Step 1: Determine the radius of the nucleus
The radius \( R \) of the nucleus is given by the formula:
\[ R = 1.3 \times A^{1/3} \times 10^{-15} \text{ m} \]
where \( A \) is the mass number.

### Step 2: Calculate the volume of the nucleus
The volume \( V \) of the nucleus can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi R^3 \]
Substituting the expression for \( R \):
\[ V = \frac{4}{3} \pi (1.3 \times A^{1/3} \times 10^{-15})^3 \]
Calculating \( R^3 \):
\[ R^3 = (1.3^3) \times (A^{1}) \times (10^{-15})^3 = 2.197 \times A \times 10^{-45} \text{ m}^3 \]
Now substituting \( R^3 \) into the volume formula:
\[ V = \frac{4}{3} \pi \times 2.197 \times A \times 10^{-45} \]
Calculating the numerical value:
\[ V \approx 9.2 \times A \times 10^{-45} \text{ m}^3 \]

### Step 3: Calculate the mass of the nucleus
The mass \( m \) of the nucleus can be calculated as:
\[ m = A \times \text{mass of one nucleon} \]
Given that the mass of one nucleon is \( 1.67 \times 10^{-27} \text{ kg} \), we have:
\[ m = A \times 1.67 \times 10^{-27} \text{ kg} \]

### Step 4: Calculate the density of the nucleus
Density \( \rho \) is given by the formula:
\[ \rho = \frac{m}{V} \]
Substituting the expressions for \( m \) and \( V \):
\[ \rho = \frac{A \times 1.67 \times 10^{-27}}{9.2 \times A \times 10^{-45}} \]
The \( A \) cancels out:
\[ \rho = \frac{1.67 \times 10^{-27}}{9.2 \times 10^{-45}} \text{ kg/m}^3 \]
Calculating the density:
\[ \rho \approx 1.8 \times 10^{17} \text{ kg/m}^3 \]

### Step 5: Calculate the number of nucleons per unit volume
The number of nucleons per unit volume \( n \) is given by:
\[ n = \frac{A}{V} \]
Substituting the expression for \( V \):
\[ n = \frac{A}{9.2 \times A \times 10^{-45}} \]
The \( A \) cancels out:
\[ n = \frac{1}{9.2 \times 10^{-45}} \text{ nucleons/m}^3 \]
Calculating the number of nucleons per unit volume:
\[ n \approx 1.09 \times 10^{44} \text{ nucleons/m}^3 \]

### Final Results
- Density of the nucleus: \( \rho \approx 1.8 \times 10^{17} \text{ kg/m}^3 \)
- Number of nucleons per unit volume: \( n \approx 1.09 \times 10^{44} \text{ nucleons/m}^3 \)