Question

If the nuclear radius of $^{125}_{52}\text{Te}$ is 6 fermi, then the nuclear radius of $^{27}_{13}\text{Al}$ in fermi is

• A. $3.6$
• B. $5$
• C. $2.5$
• D. $1.7$
• E. $4.2$

Hint:

The nuclear radius follows the $A^{1/3}$ proportionality, allowing estimation for different elements.

Answer 1

The Correct Option is A

Step 1: The nuclear radius is given by the empirical formula: \[ R = R_0 A^{1/3} \] where $A$ is the mass number and $R_0$ is a proportionality constant. 
Step 2: Using the ratio of nuclear radii: \[ \frac{R_2}{R_1} = \left( \frac{A_2}{A_1} \right)^{1/3} \] Given $R_1 = 6$ fermi for $A_1 = 125$ and $A_2 = 27$, we compute: \[ R_2 = 6 \times \left( \frac{27}{125} \right)^{1/3} \] 
Step 3: Approximating the cube root: \[ \left( \frac{27}{125} \right)^{1/3} = \frac{3}{5} = 0.6 \] 
Step 4: \[ R_2 = 6 \times 0.6 = 3.6 { fermi} \] 
Step 5: Therefore, the correct answer is (A).