Question

If the nuclear radius of $^{27}Al$ is 3.6 fermi, the nuclear radius of $^{125}Fe$ is

• A. \( 6 \times 10^{-10} \, \text{m} \)
• B. \( 6 \times 10^{-13} \, \text{m} \)
• C. \( 6 \times 10^{-15} \, \text{m} \)
• D. \( 6 \times 10^{-12} \, \text{m} \)

Hint:

Remember, the nuclear radius scales as \( A^{1/3} \). Always use the mass number to calculate the nuclear radius, and use \( R_0 \approx 1.2 \, \text{fm} \).

Answer 1

The Correct Option is C

The nuclear radius \( R \) of a nucleus is related to its mass number \( A \) by the formula: \[ R = R_0 \cdot A^{1/3} \] Where: - \( R_0 \) is a constant that can be determined experimentally (approximately \( 1.2 \, \text{fm} \) or \( 1.2 \times 10^{-15} \, \text{m} \)), - \( A \) is the mass number of the nucleus. First, we are given the radius of \(^{27}Al\), which is 3.6 fermi, and we can use the relation to find the nuclear radius of \(^{125}Fe\). We know: \[ R_{Al} = R_0 \cdot A_{Al}^{1/3} \] So, for \(^{125}Fe\): \[ R_{Fe} = R_0 \cdot A_{Fe}^{1/3} \] Substituting the given values: \[ R_{Al} = 3.6 \, \text{fm} = R_0 \cdot (27)^{1/3} \]
Thus, \[ R_0 = \frac{3.6}{(27)^{1/3}} = 1.2 \, \text{fm} \] Now, calculating the radius for \(^{125}Fe\): \[ R_{Fe} = 1.2 \cdot (125)^{1/3} \approx 6 \, \text{fm} = 6 \times 10^{-15} \, \text{m} \]
Thus, the nuclear radius of \(^{125}Fe\) is approximately \( 6 \times 10^{-15} \, \text{m} \).