Question

The ratio of volume of Al²⁷ nucleus to its surface area is (Given R₀= 1.2x10^-15 m)

• A. 2.1x10^-15 m
• B. 1.3x10^-15 m
• C. 0.22x10^-15 m
• D. 1.2x10^-15 m

Answer 1

The Correct Option is D

The volume \( V \) and surface area \( A \) of a nucleus are given by: \[ V \propto R^3, \quad A \propto R^2 \] where \( R \) is the radius of the nucleus. The ratio of volume to surface area is: \[ \frac{V}{A} \propto \frac{R^3}{R^2} = R \] Given that \( R_0 = 1.2 \times 10^{-15} \, \text{m} \), the ratio of volume to surface area is \( R_0 \). Thus, the solution is \( 1.2 \times 10^{-15} \, \text{m} \). 

Answer 2

The volume \( V \) and surface area \( A \) of a nucleus are related to its radius \( R \) as: \[ V \propto R^3 \quad \text{and} \quad A \propto R^2 \] where \( R \) is the radius of the nucleus. The ratio of the volume to surface area is then given by: \[ \frac{V}{A} \propto \frac{R^3}{R^2} = R \] Thus, the ratio of volume to surface area is directly proportional to the radius of the nucleus. \\ Given that the radius \( R \) is related to the mass number \( A \) by the formula: \[ R = R_0 A^{1/3} \] where \( R_0 \) is a constant (\( 1.2 \times 10^{-15} \, \text{m} \)) and \( A \) is the mass number of the nucleus. For \( \text{Al}^{27} \), \( A = 27 \). Thus, we can calculate \( R \) as: \[ R = 1.2 \times 10^{-15} \times 27^{1/3} \] Using the cube root of 27: \[ R = 1.2 \times 10^{-15} \times 3 = 3.6 \times 10^{-15} \, \text{m} \] Therefore, the ratio of volume to surface area is simply \( R \), which is \( 1.2 \times 10^{-15} \, \text{m} \).

Thus, the correct ratio of volume to surface area is \( 1.2 \times 10^{-15} \, \text{m} \).