The ratio of the radii of the nuclei `._(13)^(27)Al " and " ._(52)Te^(125)` is approximately-

Options

A. `6 : 10`
B. `13 : 52`
C. `40 : 177`
D. `14 : 73`

Answer

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

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

where:
- \(R\) is the radius of the nucleus,
- \(R_0\) is a constant approximately equal to \(1.25 \times 10^{-15}\) meters,
- \(A\) is the mass number of the nucleus.

### Step 1: Calculate the radius of aluminum nucleus \(R_1\)

For aluminum, the mass number \(A_1 = 27\). Therefore, we can calculate \(R_1\) as follows:

\[ R_1 = R_0 \cdot A_1^{1/3} = 1.25 \times 10^{-15} \cdot 27^{1/3} \]

### Step 2: Calculate the radius of tellurium nucleus \(R_2\)

For tellurium, the mass number \(A_2 = 125\). We can calculate \(R_2\) as follows:

\[ R_2 = R_0 \cdot A_2^{1/3} = 1.25 \times 10^{-15} \cdot 125^{1/3} \]

### Step 3: Find the ratio of the radii \(R_1\) to \(R_2\)

Now, we need to find the ratio of the two radii:

\[ \frac{R_1}{R_2} = \frac{1.25 \times 10^{-15} \cdot 27^{1/3}}{1.25 \times 10^{-15} \cdot 125^{1/3}} \]

The \(1.25 \times 10^{-15}\) cancels out:

\[ \frac{R_1}{R_2} = \frac{27^{1/3}}{125^{1/3}} = \left(\frac{27}{125}\right)^{1/3} \]

### Step 4: Simplify the ratio

We can simplify \(\frac{27}{125}\):

\[ \frac{27}{125} = \frac{3^3}{5^3} = \left(\frac{3}{5}\right)^3 \]

Thus, we have:

\[ \frac{R_1}{R_2} = \left(\frac{3}{5}\right) = \frac{3}{5} \]

### Conclusion

The ratio of the radii of the nuclei of aluminum and tellurium is approximately:

\[ \frac{R_1}{R_2} \approx \frac{3}{5} \]