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. `6.0 fm`
B. `9.6 fm`
C. `12.0 fm`
D. `4.8 fm`

Answer

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

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

where \( R_0 \) is a constant (approximately \( 1.2 \) to \( 1.3 \) Fermi) and \( A \) is the mass number.

### Step-by-Step Solution:

1. **Identify the mass numbers**:
- For Aluminum (\( _{13}^{27}\text{Al} \)), the mass number \( A_1 = 27 \).
- For Tellurium (\( _{52}^{125}\text{Te} \)), the mass number \( A_2 = 125 \).

2. **Write the equation for the radius of Aluminum**:
\[ R_1 = R_0 A_1^{1/3} = R_0 \cdot 27^{1/3} \]
Given that \( R_1 = 3.6 \) Fermi, we can express this as:
\[ 3.6 = R_0 \cdot 27^{1/3} \]

3. **Write the equation for the radius of Tellurium**:
\[ R_2 = R_0 A_2^{1/3} = R_0 \cdot 125^{1/3} \]

4. **Divide the equations for the two nuclei**:
\[ \frac{R_1}{R_2} = \frac{R_0 \cdot 27^{1/3}}{R_0 \cdot 125^{1/3}} \]
The \( R_0 \) cancels out:
\[ \frac{3.6}{R_2} = \frac{27^{1/3}}{125^{1/3}} \]

5. **Simplify the right-hand side**:
\[ \frac{3.6}{R_2} = \left(\frac{27}{125}\right)^{1/3} \]

6. **Calculate \( \frac{27}{125} \)**:
\[ \frac{27}{125} = \frac{3^3}{5^3} = \left(\frac{3}{5}\right)^3 \]
Therefore:
\[ \left(\frac{27}{125}\right)^{1/3} = \frac{3}{5} \]

7. **Substitute back into the equation**:
\[ \frac{3.6}{R_2} = \frac{3}{5} \]

8. **Cross-multiply to solve for \( R_2 \)**:
\[ 3.6 \cdot 5 = 3 \cdot R_2 \]
\[ 18 = 3 \cdot R_2 \]
\[ R_2 = \frac{18}{3} = 6 \text{ Fermi} \]

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

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