Question

If the radius of a nucleus having \(13\) protons and \(14\) neutrons is \(3.6\,\text{fm}\), then the ratio of the volume to surface area of a nucleus having \(53\) protons and \(72\) neutrons is

• A. \(12\,\text{fm}\)
• B. \(3\,\text{fm}\)
• C. \(6\,\text{fm}\)
• D. \(2\,\text{fm}\) \bigskip

Hint:

For a spherical nucleus, \[ R=R_0A^{1/3}. \] Also, \[ \frac{\text{Volume}}{\text{Surface Area}} = \frac{R}{3}. \] So once the nuclear radius is known, the required ratio can be obtained directly.

Answer 1

The Correct Option is D

Concept: Nuclear radius \(R = R_0 A^{1/3}\). Volume/Surface area = \(R/3\). Find \(R_0\) from first nucleus, apply to second.

Step 1: \(A_1=27, R_1=3.6\) fm \(\Rightarrow 3.6 = R_0(27)^{1/3} \Rightarrow R_0 = 1.2\) fm. \(A_2=125, R_2 = 1.2\times5 = 6\) fm. \(V/S = 6/3 = 2\) fm.

Step 2: Write the final answer. \(\boxed{\frac{V}{S}=2\,\text{fm}}\)