Assume that the mass of a nucleus is approximately given by `M=Am_p`where A is the mass number.Estimate the density of matter in `kgm^-3`inside a nucleus.

Explanation

To estimate the density of matter inside a nucleus, we can follow these steps:

### Step 1: Understand the mass of the nucleus
The mass \( M \) of the nucleus can be approximated as:
\[ M = A \cdot m_p \]
where \( A \) is the mass number (the total number of protons and neutrons in the nucleus) and \( m_p \) is the mass of a proton.

### Step 2: Determine the volume of the nucleus
The volume \( V \) of the nucleus can be modeled as a sphere. The formula for the volume of a sphere is:
\[ V = \frac{4}{3} \pi r^3 \]
In nuclear physics, the radius \( r \) of the nucleus can be estimated using the formula:
\[ r = r_0 A^{1/3} \]
where \( r_0 \) is a constant approximately equal to \( 1.2 \times 10^{-15} \) meters.

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

### Step 4: Calculate the density
Density \( D \) is defined as mass per unit volume:
\[ D = \frac{M}{V} \]
Substituting the expressions for \( M \) and \( V \):
\[ D = \frac{A \cdot m_p}{\frac{4}{3} \pi r_0^3 A} = \frac{m_p}{\frac{4}{3} \pi r_0^3} \]
Notice that \( A \) cancels out.

### Step 5: Substitute known values
The mass of a proton \( m_p \) is approximately \( 1.67 \times 10^{-27} \) kg. The value of \( r_0 \) is approximately \( 1.2 \times 10^{-15} \) m. Now, substituting these values into the density formula:
\[ D = \frac{1.67 \times 10^{-27}}{\frac{4}{3} \pi (1.2 \times 10^{-15})^3} \]

### Step 6: Calculate the volume term
Calculating the volume term:
\[ \frac{4}{3} \pi (1.2 \times 10^{-15})^3 \approx \frac{4}{3} \times 3.14 \times (1.728 \times 10^{-45}) \approx 7.238 \times 10^{-45} \text{ m}^3 \]

### Step 7: Final density calculation
Now substituting this back into the density equation:
\[ D \approx \frac{1.67 \times 10^{-27}}{7.238 \times 10^{-45}} \approx 2.31 \times 10^{17} \text{ kg/m}^3 \]

### Final Answer
Thus, the estimated density of matter inside a nucleus is approximately:
\[ D \approx 2.31 \times 10^{17} \text{ kg/m}^3 \]