Find the binding energy of a nucleus consisting of equal numbers of protons and neutrons and having the radius one and a half time smaller than of `Al^(27)` nucleus.

Explanation

To find the binding energy of a nucleus consisting of equal numbers of protons and neutrons and having a radius one and a half times smaller than that of the Al-27 nucleus, we can follow these steps:

### Step 1: Determine the radius of the Al-27 nucleus
The radius of a nucleus can be estimated using the formula:
\[ R = R_0 A^{1/3} \]
where \( R_0 \) is a constant (approximately \( 1.2 \, \text{fm} \)) and \( A \) is the mass number. For Al-27, \( A = 27 \).

### Step 2: Calculate the radius of the Al-27 nucleus
Using the formula:
\[ R_{Al} = R_0 (27)^{1/3} \]
Substituting \( R_0 = 1.2 \, \text{fm} \):
\[ R_{Al} = 1.2 \times (27)^{1/3} \]

Calculating \( (27)^{1/3} \):
\[ (27)^{1/3} = 3 \]
Thus,
\[ R_{Al} = 1.2 \times 3 = 3.6 \, \text{fm} \]

### Step 3: Find the radius of the new nucleus
Since the radius of the new nucleus is one and a half times smaller than that of Al-27:
\[ R_{new} = \frac{R_{Al}}{1.5} = \frac{3.6 \, \text{fm}}{1.5} = 2.4 \, \text{fm} \]

### Step 4: Calculate the mass number of the new nucleus
Using the radius formula again:
\[ R_{new} = R_0 A^{1/3} \]
We can rearrange this to find \( A \):
\[ A = \left(\frac{R_{new}}{R_0}\right)^3 \]
Substituting \( R_{new} = 2.4 \, \text{fm} \) and \( R_0 = 1.2 \, \text{fm} \):
\[ A = \left(\frac{2.4}{1.2}\right)^3 = (2)^3 = 8 \]

### Step 5: Determine the number of protons and neutrons
Since the nucleus consists of equal numbers of protons and neutrons:
- Number of protons \( Z = \frac{A}{2} = \frac{8}{2} = 4 \)
- Number of neutrons \( N = \frac{A}{2} = 4 \)

### Step 6: Calculate the binding energy
The binding energy \( E_b \) can be calculated using the semi-empirical mass formula:
\[ E_b = Z \cdot m_p c^2 + N \cdot m_n c^2 - m_{nucleus} c^2 \]
where:
- \( m_p \) = mass of a proton \( \approx 0.00867 \, \text{amu} \)
- \( m_n \) = mass of a neutron \( \approx 0.00783 \, \text{amu} \)
- \( m_{nucleus} \) = mass of the nucleus \( \approx 0.00531 \, \text{amu} \)

Substituting the values:
\[ E_b = 4 \cdot 0.00867 + 4 \cdot 0.00783 - 0.00531 \]
Calculating:
\[ E_b = 0.03468 + 0.03132 - 0.00531 \]
\[ E_b = 0.06069 \, \text{amu} \cdot c^2 \]

### Step 7: Convert binding energy to MeV
Using the conversion \( 1 \, \text{amu} \cdot c^2 \approx 931.5 \, \text{MeV} \):
\[ E_b = 0.06069 \times 931.5 \approx 56.5 \, \text{MeV} \]

### Final Answer
The binding energy of the nucleus is approximately \( 56.5 \, \text{MeV} \).