If radius of the `._(13)^(27)Al` nucleus is taken to be `R_(AI)`, then the radius of `._(53)^(125)Te` nucleus is nearly

Options

A. `(53/13)^(1/3)R_(Al)`
B. `5/3 R_(Al)`
C. `3/5 R_(Al)`
D. `(13/53)^(1/3)R_(Al)`

Answer

To find the radius of the \( _{53}^{125}Te \) nucleus in relation to the radius of the \( _{13}^{27}Al \) nucleus, we can use the formula that relates the radius of a nucleus to its mass number \( A \):

### Step 1: Understand the relationship between radius and mass number
The radius \( R \) of a nucleus is given by the formula:
\[ R \propto A^{1/3} \]
where \( A \) is the mass number of the nucleus.

### Step 2: Write the expressions for the radii of both nuclei
Let \( R_{Al} \) be the radius of the aluminum nucleus and \( R_{Te} \) be the radius of the tellurium nucleus. We can express the radii in terms of their mass numbers:
\[ R_{Al} \propto (27)^{1/3} \]
\[ R_{Te} \propto (125)^{1/3} \]

### Step 3: Set up the ratio of the radii
To find the ratio of the radii, we can write:
\[ \frac{R_{Al}}{R_{Te}} = \frac{(27)^{1/3}}{(125)^{1/3}} = \left(\frac{27}{125}\right)^{1/3} \]

### Step 4: Simplify the ratio
Calculating \( \frac{27}{125} \):
\[ \frac{27}{125} = \left(\frac{3}{5}\right)^3 \]
Thus,
\[ \left(\frac{27}{125}\right)^{1/3} = \frac{3}{5} \]

### Step 5: Relate the radii
From the ratio we derived:
\[ \frac{R_{Al}}{R_{Te}} = \frac{3}{5} \]
This implies:
\[ R_{Al} = \frac{3}{5} R_{Te} \]

### Step 6: Solve for \( R_{Te} \)
Rearranging gives:
\[ R_{Te} = \frac{5}{3} R_{Al} \]

### Conclusion
Thus, the radius of the \( _{53}^{125}Te \) nucleus is nearly:
\[ R_{Te} \approx \frac{5}{3} R_{Al} \]