Nucleus: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Nucleus. Understanding the underlying formula is key to arriving at the correct answer below:

Change in mass number $\Delta A = 200 - 168 = 32$. Since each $\alpha$ particle reduces $A$ by 4, number of $\alpha$ particles = $32 / 4 = 8$. Expected $Z$ after 8 $\alpha$ emissions = $90 - 8(2) = 74$. Actual final $Z$ is 80. The increase in $Z$ by 6 requires the emission of 6 $\beta$ particles.