Q131 / 180 · COMEDK 2024 (Evening Shift) · Physics
$\text { If the nuclear radius of }{ }^{27} \mathrm{Al} \text { is } 3.6 \text { fermi, the nuclear radius of }{ }^{125} \mathrm{Fe} \text { is }$
- A$6 \times 10^{-12} \mathrm{~m}$
- B$6 \times 10^{-10} \mathrm{~m}$
- C$6 \times 10^{-13} \mathrm{~m}$
- Correct answerD$6 \times 10^{-15} \mathrm{~m}$
Solution
The nuclear radius $R$ is given by the formula $R = R_0 A^{1/3}$, where $R_0$ is a constant and $A$ is the mass number of the nucleus.
For $^{27}\mathrm{Al}$, the mass number $A_1 = 27$ and the radius $R_1 = 3.6 \text{ fermi}$.
Using the formula, $3.6 = R_0 (27)^{1/3} = R_0 \times 3$, which gives $R_0 = 1.2 \text{ fermi}$.
For $^{125}\mathrm{Fe}$, the mass number $A_2 = 125$. The radius $R_2$ is given by $R_2 = R_0 (125)^{1/3} = 1.2 \times 5 = 6.0 \text{ fermi}$.
Since $1 \text{ fermi} = 10^{-15} \text{ m}$, the radius $R_2 = 6.0 \times 10^{-15} \text{ m}$.
Answer: $6 \times 10^{-15} \mathrm{~m}$
For $^{27}\mathrm{Al}$, the mass number $A_1 = 27$ and the radius $R_1 = 3.6 \text{ fermi}$.
Using the formula, $3.6 = R_0 (27)^{1/3} = R_0 \times 3$, which gives $R_0 = 1.2 \text{ fermi}$.
For $^{125}\mathrm{Fe}$, the mass number $A_2 = 125$. The radius $R_2$ is given by $R_2 = R_0 (125)^{1/3} = 1.2 \times 5 = 6.0 \text{ fermi}$.
Since $1 \text{ fermi} = 10^{-15} \text{ m}$, the radius $R_2 = 6.0 \times 10^{-15} \text{ m}$.
Answer: $6 \times 10^{-15} \mathrm{~m}$
Hence, the correct answer is option: 'D'.