If the nuclear of radius of `O^8` is `2.4` fermi, then radius of `Al^27` would be

Options

A. 6.0 fermi
B. 5.6 fermi
C. 3.6 fermi
D. 3.0 fermi

Answer

To find the radius of the aluminum nucleus \( Al^{27} \) given the radius of the oxygen nucleus \( O^{8} \), we can use the formula for the radius of a nucleus:

\[ R = R_0 \cdot A^{1/3} \]

where:
- \( R \) is the radius of the nucleus,
- \( R_0 \) is a constant (approximately \( 1.2 \) to \( 1.3 \) fermi),
- \( A \) is the mass number of the nucleus.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- The radius of the oxygen nucleus \( R_O = 2.4 \) fermi.
- The mass number of oxygen \( A_O = 8 \).
- The mass number of aluminum \( A_{Al} = 27 \).

2. **Set Up the Ratio of Radii**:
The radii of the nuclei can be related through their mass numbers:
\[ \frac{R_O}{R_{Al}} = \left( \frac{A_O}{A_{Al}} \right)^{1/3} \]

3. **Substitute the Known Values**:
\[ \frac{2.4}{R_{Al}} = \left( \frac{8}{27} \right)^{1/3} \]

4. **Calculate the Mass Number Ratio**:
Calculate \( \frac{8}{27} \):
\[ \frac{8}{27} = \frac{2^3}{3^3} = \left( \frac{2}{3} \right)^3 \]
Therefore,
\[ \left( \frac{8}{27} \right)^{1/3} = \frac{2}{3} \]

5. **Set Up the Equation**:
Now substituting back into the equation:
\[ \frac{2.4}{R_{Al}} = \frac{2}{3} \]

6. **Cross Multiply to Solve for \( R_{Al} \)**:
\[ 2.4 = R_{Al} \cdot \frac{2}{3} \]
Rearranging gives:
\[ R_{Al} = 2.4 \cdot \frac{3}{2} \]

7. **Calculate \( R_{Al} \)**:
\[ R_{Al} = 2.4 \cdot 1.5 = 3.6 \text{ fermi} \]

### Conclusion:
The radius of the aluminum nucleus \( Al^{27} \) is \( 3.6 \) fermi.