Question

A nucleus with atomic mass number \( A \) produces another nucleus by losing 2 alpha particles. The volume of the new nucleus is 60 times that of the alpha particle. The atomic mass number \( A \) of the original nucleus is:

• A. \( 228 \)
• B. \( 238 \)
• C. \( 248 \)
• D. \( 244 \)

Hint:

In nuclear physics, the volume of a nucleus is proportional to the cube of its atomic mass number (\( V \propto A^{3} \)). Use this relation to determine mass numbers when nuclear transformations occur.

Answer 1

The Correct Option is C

To determine the atomic mass number \( A \) of the original nucleus, we need to consider the effect of losing two alpha particles on the original nucleus. Each alpha particle consists of 2 protons and 2 neutrons, which means it has an atomic mass number of 4.

Thus, when a nucleus loses two alpha particles, the decrease in the atomic mass number is:

\( 2 \times 4 = 8 \)

Let's denote the atomic mass number of the original nucleus as \( A \). After losing two alpha particles, the new atomic mass number becomes \( A - 8 \).

The problem states that the volume of the new nucleus is 60 times that of an alpha particle. Since the volume of a nucleus is proportional to its atomic mass number, we can express this relationship as:

\( A - 8 = 60 \times 4 \)

Solving for \( A \):

\( A - 8 = 240 \)

\( A = 240 + 8 = 248 \)

Thus, the atomic mass number \( A \) of the original nucleus is 248.

Answer 2

Step 1: Understanding the Problem When a nucleus loses 2 alpha particles, each alpha particle has a mass number of 4. So the total mass lost by the nucleus is: \[ 2 \times 4 = 8 \] Let the atomic mass of the original nucleus be \( A \). The remaining nucleus will have an atomic mass of: \[ A' = A - 8 \] Step 2: Volume Relation and Mass Number The volume of a nucleus is proportional to the cube of the atomic mass number (\( V \propto A^{3} \)). Given that the volume of the new nucleus is 60 times that of an alpha particle, we set up the equation: \[ A'^{3} = 60 \times (4)^{3} \] Since \( (4)^3 = 64 \), we get: \[ A'^{3} = 60 \times 64 = 3840 \] \[ A' = \sqrt[3]{3840} \] Approximating: \[ A' \approx 240 \] Since \( A' = A - 8 \), we substitute: \[ A = 240 + 8 = 248 \] Step 3: Conclusion Thus, the atomic mass number of the original nucleus is \( 248 \).