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

To determine the radius of a nucleus given its mass number, we use the empirical formula for the nuclear radius:

\(R = R_0 \cdot A^{1/3}\)

where:

• \(R_0\) is a constant approximately equal to 1.2 femtometers (fm).
• \(A\) is the mass number of the nucleus.

Given in the question, the mass number \(A\) is 125. Let's calculate the radius:

\(R = 1.2 \, \text{fm} \times 125^{1/3}\)

First, we find \(125^{1/3}\):

\(125^{1/3} = 5\) (since 5³ = 125).

Substituting this value back into the formula, we get:

\(R = 1.2 \times 5 = 6.0 \, \text{fm}\)

Therefore, the radius of the nucleus with mass number 125 is 6.0 fm, which matches option 1: "6.0 fm".

Let's verify by ruling out other options:

• Option 2: 30 fm would require a mass number significantly larger than 125.
• Option 3: 72 fm is extraordinarily large, suggesting a much larger nucleus (incorrect for A = 125).
• Option 4: 150 fm is unrealistically high, suggesting an enormous mass number, which is not the case here.

Thus, the correct answer is confirmed to be: \(6.0 \, \text{fm}\).

Answer 2

Rather than only computing the exact value, this method estimates the expected order of magnitude first and then checks which listed option actually fits, which is a useful cross-check for numeric answers.

The nuclear radius formula \( R = R_0 A^{1/3} \), with \( R_0 \approx 1.2\,\text{fm} \), tells us that \( R \) grows only with the cube root of the mass number, not with the mass number itself. For \( A = 125 \), the cube root is a modest number, since \( 5^3 = 125 \) exactly, \( A^{1/3} = 5 \).

So the expected radius should be roughly \( 1.2\,\text{fm} \times 5 \), a single-digit multiple of a femtometre, not tens or hundreds of femtometres. With this expectation, we can check the listed options:

30 fm would require \( A^{1/3} \approx 25 \), meaning \( A \approx 15{,}625 \), far larger than 125, so this is much too big.
72 fm would require an even larger, unrealistic mass number, ruling it out further.
150 fm is larger still, corresponding to an enormous, physically implausible nucleus for \( A = 125 \).
6.0 fm matches the direct computation \( 1.2 \times 5 = 6.0\,\text{fm} \) exactly, and sits comfortably in the expected single-digit femtometre range for a nucleus of this size.

Only the smallest of the four options is consistent with how modestly the radius grows for a mass number of 125.

Therefore, the correct answer is 6.0 fm.