Question

If the ratio of the mass numbers of two nuclei is 27:125, then the ratio of their surface areas is

• A. 3:5
• B. 9:25
• C. 27:125
• D. 1:1

Hint:

For nuclei, remember these proportionality relationships with the mass number A: - Radius \(R \propto A^{1/3}\) - Surface Area \(S \propto R^2 \propto (A^{1/3})^2 \propto A^{2/3}\) - Volume \(V \propto R^3 \propto (A^{1/3})^3 \propto A\)

Answer 1

The Correct Option is B

The radius (R) of a nucleus is empirically found to be related to its mass number (A) by the formula:
\( R = R_0 A^{1/3} \), where \(R_0\) is a constant approximately equal to 1.2 fm.
This relationship implies that the volume of a nucleus is proportional to its mass number, meaning nuclear density is roughly constant.
We are given the ratio of the mass numbers of two nuclei:
\( \frac{A_1}{A_2} = \frac{27}{125} \).
Let's find the ratio of their radii.
\( \frac{R_1}{R_2} = \frac{R_0 A_1^{1/3}}{R_0 A_2^{1/3}} = \left(\frac{A_1}{A_2}\right)^{1/3} \).
\( \frac{R_1}{R_2} = \left(\frac{27}{125}\right)^{1/3} = \frac{\sqrt[3]{27}}{\sqrt[3]{125}} = \frac{3}{5} \).
Now, we need to find the ratio of their surface areas.
The surface area of a nucleus is assumed to be spherical, so the formula is \( S = 4\pi R^2 \).
The ratio of the surface areas is:
\( \frac{S_1}{S_2} = \frac{4\pi R_1^2}{4\pi R_2^2} = \left(\frac{R_1}{R_2}\right)^2 \).
Substitute the ratio of the radii we found.
\( \frac{S_1}{S_2} = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \).
The ratio of their surface areas is 9:25.