Question

If the radius of the nucleus of \(^{13} \text{Al}^{27}\) is 3.6 fermi, then the radius of the nucleus of \(^{52} \text{Te}^{125}\) is: 
 

• A. \(8 \, \text{fermi}\)
• B. \(6 \, \text{fermi}\)
• C. \(5 \, \text{fermi}\)
• D. \(4 \, \text{fermi}\)

Hint:

The nuclear radius is proportional to \( A^{1/3} \), where \(A\) is the mass number.

Answer 1

The Correct Option is B

The radius of a nucleus is proportional to the cube root of its mass number (\(A\)): \[ R \propto A^{1/3}. \] For \(^{13} \text{Al}^{27}\), \[ R_1 = 3.6 \, \text{fermi}, \quad A_1 = 27, \quad A_2 = 125. \] \[ \frac{R_2}{R_1} = \left(\frac{A_2}{A_1}\right)^{1/3} \quad \Rightarrow \quad R_2 = R_1 \cdot \left(\frac{125}{27}\right)^{1/3}. \] \[ R_2 = 3.6 \cdot 2 \quad \Rightarrow \quad R_2 = 6 \, \text{fermi}. \]