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 \( 248 \).