If the nucleus of `._13Al^27` has a nuclear radius of about 3.6 fm, then `._52Te^125` would have its radius approximately as

Options

A. 9.6 fm
B. 12 fm
C. 4.8 fm
D. 6 fm

Explanation

To find the nuclear radius of the isotope \( _{52}^{125}\text{Te} \) (Tellurium) given that the nuclear radius of \( _{13}^{27}\text{Al} \) (Aluminum) is approximately 3.6 fm, we can use the formula for the radius of a nucleus, which is given by:

\[ r \propto A^{1/3} \]

where \( r \) is the radius and \( A \) is the mass number of the nucleus.

### Step 1: Write the relationship between the radii of the two nuclei

Using the proportionality, we can express the radii of Aluminum and Tellurium as:

\[ \frac{r_{\text{Al}}}{r_{\text{Te}}} = \frac{A_{\text{Al}}^{1/3}}{A_{\text{Te}}^{1/3}} \]

### Step 2: Substitute the known values

The mass number of Aluminum \( A_{\text{Al}} = 27 \) and the mass number of Tellurium \( A_{\text{Te}} = 125 \). Therefore, we can write:

\[ \frac{r_{\text{Al}}}{r_{\text{Te}}} = \frac{27^{1/3}}{125^{1/3}} \]

### Step 3: Calculate the cube roots

Now, we calculate the cube roots:

- The cube root of 27 is \( 3 \) (since \( 3^3 = 27 \)).
- The cube root of 125 is \( 5 \) (since \( 5^3 = 125 \)).

Thus, we have:

\[ \frac{r_{\text{Al}}}{r_{\text{Te}}} = \frac{3}{5} \]

### Step 4: Express \( r_{\text{Te}} \) in terms of \( r_{\text{Al}} \)

Rearranging the equation gives us:

\[ r_{\text{Te}} = r_{\text{Al}} \cdot \frac{5}{3} \]

### Step 5: Substitute the known radius of Aluminum

Given that \( r_{\text{Al}} = 3.6 \, \text{fm} \):

\[ r_{\text{Te}} = 3.6 \, \text{fm} \cdot \frac{5}{3} \]

### Step 6: Calculate \( r_{\text{Te}} \)

Now we perform the multiplication:

\[ r_{\text{Te}} = 3.6 \cdot \frac{5}{3} = 3.6 \cdot 1.6667 = 6 \, \text{fm} \]

Thus, the nuclear radius of \( _{52}^{125}\text{Te} \) is approximately \( 6 \, \text{fm} \).

### Final Answer

The radius of \( _{52}^{125}\text{Te} \) is approximately \( 6 \, \text{fm} \).

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