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

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

where \( R_0 \) is a constant (approximately \( 1.2 \, \text{fm} \)) and \( A \) is the mass number of the nucleus.

### Step 1: Calculate the radius of the aluminum nucleus
The mass number \( A \) for aluminum \( _{13}^{27}Al \) is 27. Thus, the radius \( R_{Al} \) can be calculated as:

\[ R_{Al} = R_0 \cdot (27)^{1/3} \]

### Step 2: Calculate the radius of the tellurium nucleus
The mass number \( A \) for tellurium \( _{53}^{125}Te \) is 125. Therefore, the radius \( R_{Te} \) can be calculated as:

\[ R_{Te} = R_0 \cdot (125)^{1/3} \]

### Step 3: Relate the two radii
To find the ratio of the two radii, we can express \( R_{Te} \) in terms of \( R_{Al} \):

\[ \frac{R_{Te}}{R_{Al}} = \frac{R_0 \cdot (125)^{1/3}}{R_0 \cdot (27)^{1/3}} = \frac{(125)^{1/3}}{(27)^{1/3}} = \left(\frac{125}{27}\right)^{1/3} \]

### Step 4: Calculate the numerical values
Now we can calculate \( \frac{125}{27} \):

\[ \frac{125}{27} \approx 4.62963 \]

Now, we take the cube root of this value:

\[ R_{Te} = R_{Al} \cdot (4.62963)^{1/3} \]

Calculating the cube root:

\[ (4.62963)^{1/3} \approx 1.669 \]

Thus,

\[ R_{Te} \approx R_{Al} \cdot 1.669 \]

### Final Result
If the radius of the aluminum nucleus is known, the radius of the tellurium nucleus can be approximated as:

\[ R_{Te} \approx 1.669 \cdot R_{Al} \]