Question

If the volume of nucleus having mass number 2 is V, then that for the nucleus having mass number 8 is

• A. 2V
• B. 3V
• C. 4V
• D. 6V
• E. 9V

Hint:

Nuclear density is constant because volume is proportional to mass number.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The volume of an atomic nucleus is related to its mass number (\(A\)), which is the total number of protons and neutrons. Experiments show that nuclear matter has a nearly constant density, which implies a direct relationship between the volume and the number of nucleons.
Step 2: Key Formula or Approach:
The radius \(R\) of a nucleus is given by the empirical formula:
\[ R = R_0 A^{1/3} \] where \(R_0\) is a constant (\(\approx 1.2 \times 10^{-15}\) m) and \(A\) is the mass number.
Assuming the nucleus is spherical, its volume \(V_{\text{nucleus}}\) is:
\[ V_{\text{nucleus}} = \frac{4}{3}\pi R^3 \] By substituting the formula for \(R\), we can find the relationship between volume and mass number.
Step 2: Detailed Explanation:
Let's derive the relationship between volume (\(V_{\text{nucleus}}\)) and mass number (\(A\)):
\[ V_{\text{nucleus}} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 \] \[ V_{\text{nucleus}} = \frac{4}{3}\pi R_0^3 (A^{1/3})^3 \] \[ V_{\text{nucleus}} = \left(\frac{4}{3}\pi R_0^3\right) A \] Since \(\frac{4}{3}\pi R_0^3\) is a constant, we can see that the volume of a nucleus is directly proportional to its mass number:
\[ V_{\text{nucleus}} \propto A \] We can use this proportionality to solve the problem. Let \(V_1\) be the volume for mass number \(A_1\), and \(V_2\) be the volume for mass number \(A_2\).
\[ \frac{V_2}{V_1} = \frac{A_2}{A_1} \] Given:
- For mass number \(A_1 = 2\), the volume is \(V_1 = V\).
- We need to find the volume \(V_2\) for mass number \(A_2 = 8\).
Substitute the values into the ratio:
\[ \frac{V_2}{V} = \frac{8}{2} \] \[ \frac{V_2}{V} = 4 \] \[ V_2 = 4V \] Step 3: Final Answer:
The volume of the nucleus with mass number 8 is 4V, which corresponds to option (C).