Question

The radius of a nucleus of mass number 125 is:

• A. 6.0 fm
• B. 30 fm
• C. 72 fm
• D. 150 fm

Hint:

Use \( R = 1.3 A^{1/3} \, \text{fm} \). Remember: \( 125^{1/3} = 5 \) (perfect cube → quick calculation).

Answer 1

The Correct Option is A

Concept: The radius of a nucleus is given by the empirical formula: \[ R = R_0 A^{1/3} \] Where:

\( R_0 \approx 1.3 \, \text{fm} \)
\( A \) = mass number

Step 1: Substitute the given mass number. \[ A = 125 \] \[ R = 1.3 \times 125^{1/3} \]
Step 2: Evaluate cube root. \[ 125^{1/3} = 5 \]
Step 3: Calculate radius. \[ R = 1.3 \times 5 = 6.5 \, \text{fm} \] Using the commonly accepted approximation \( R_0 \approx 1.2 \, \text{fm} \): \[ R = 1.2 \times 5 = 6.0 \, \text{fm} \]
Step 4: Choose the closest option. The nearest value is \( 6.0 \, \text{fm} \).

Answer 2

Instead of substituting directly into \( R = R_0 A^{1/3} \), this approach uses a ratio comparison against a convenient reference nucleus whose radius is easy to state, scaling from there.

Take a small reference nucleus with mass number \( A_{\text{ref}} = 1 \) (a single nucleon), whose "radius" by the same formula is just \( R_0 \approx 1.2\,\text{fm} \), the accepted empirical constant.

Since the nuclear radius scales with the cube root of the mass number, the radius of any other nucleus relates to this reference by:

\[ \frac{R}{R_{\text{ref}}} = \left(\frac{A}{A_{\text{ref}}}\right)^{1/3} = A^{1/3} \]

For the nucleus in question, \( A = 125 \). Since \( 5 \times 5 \times 5 = 125 \), the cube root of 125 is exactly 5. So the radius scales up by a clean factor of 5 relative to the single-nucleon reference:

\[ R = 5 \times R_0 = 5 \times 1.2\,\text{fm} = 6.0\,\text{fm} \]

This scaling approach reaches the same figure without needing to treat \( R = R_0 A^{1/3} \) as one combined substitution, it separates the constant \( R_0 \) from the scaling factor \( A^{1/3} \) so each part of the answer's origin is clear.

Therefore, the radius of the nucleus is 6.0 fm.