Question

The ratio of the surface area of the nuclei \(_{52}\text{Te}^{125}\) to that of \(_{13}\text{Al}^{27}\) is

• A. \(\frac{5}{3}\)
• B. \(\frac{125}{17}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{25}{9}\)
• E. \(\frac{3}{5}\)

Hint:

Nuclear physics ratios usually involve cube roots ($1/3$ for Radius) or mass numbers directly ($1$ for Volume). Surface area is the "middle ground" using the power of $2/3$.

Answer 1

The Correct Option is D

Concept: The size of a nucleus is related to its mass number (\(A\)).
• Nuclear Radius (\(R\)): The radius is given by \(R = R_0 A^{1/3}\), where \(R_0\) is a constant.
• Surface Area (\(S\)): Assuming the nucleus is spherical, surface area \(S = 4\pi R^2\).
• Area Proportionality: Substituting the radius formula, \(S \propto (A^{1/3})^2 \implies S \propto A^{2/3}\).

Step 1: Identify mass numbers and set up the ratio.
Given: - For Tellurium (\(\text{Te}\)): \(A_1 = 125\) - For Aluminum (\(\text{Al}\)): \(A_2 = 27\) The ratio of their surface areas is: \[ \frac{S_{Te}}{S_{Al}} = \left( \frac{A_1}{A_2} \right)^{2/3} \]

Step 2: Perform the mathematical simplification.
\[ \frac{S_{Te}}{S_{Al}} = \left( \frac{125}{27} \right)^{2/3} \] Calculate the cube root first (\(1/3\)): \[ \left( \frac{125}{27} \right)^{1/3} = \frac{5}{3} \] Now square the result (\(2\)): \[ \left( \frac{5}{3} \right)^2 = \frac{25}{9} \]