If radius of the `._(13)^(27)Al` nucleus is estimated to be `3.6` Fermi, then the radius of `._(52)^(125)Te` nucleus be nerarly:

Options

A. `6` Fermi
B. `8` Fermi
C. `4` Fermi
D. `5` Fermi

Explanation

To find the radius of the \( _{52}^{125}Te \) nucleus given the radius of the \( _{13}^{27}Al \) nucleus, we can use the empirical formula for the radius of a nucleus, which is given by:

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

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 mass numbers**:
- For \( _{13}^{27}Al \), the mass number \( A_{Al} = 27 \).
- For \( _{52}^{125}Te \), the mass number \( A_{Te} = 125 \).

2. **Use the radius formula for both nuclei**:
- The radius of the aluminium nucleus is given as \( R_{Al} = 3.6 \) Fermi.
- The radius of the tantalum nucleus can be expressed as:
\[ R_{Te} = R_0 A_{Te}^{1/3} \]
- Similarly, for aluminium:
\[ R_{Al} = R_0 A_{Al}^{1/3} \]

3. **Set up the ratio of the radii**:
- We can express the ratio of the radii as:
\[ \frac{R_{Te}}{R_{Al}} = \frac{A_{Te}^{1/3}}{A_{Al}^{1/3}} \]

4. **Substitute the mass numbers**:
- Plugging in the values:
\[ \frac{R_{Te}}{3.6} = \left(\frac{125}{27}\right)^{1/3} \]

5. **Calculate the ratio**:
- First, calculate \( \frac{125}{27} \):
\[ \frac{125}{27} \approx 4.62963 \]
- Now take the cube root:
\[ \left(4.62963\right)^{1/3} \approx 1.669 \]

6. **Calculate the radius of \( _{52}^{125}Te \)**:
- Now substitute back to find \( R_{Te} \):
\[ R_{Te} = 3.6 \times 1.669 \approx 6.02 \text{ Fermi} \]

### Final Answer:
The radius of the \( _{52}^{125}Te \) nucleus is approximately \( 6 \) Fermi.