Question

Consider the following radioactive decay process

Hint:

$\alpha$-decay reduces atomic number by 2 and mass number by 4, while $\beta^-$-decay increases atomic number by 1

Answer 1

Correct Answer: 4

Step 1: Apply mass number conservation.
\[ 238 = 210 + 7 \times 4 \]
\[ 238 = 210 + 28 = 238 \]
Mass number is balanced, so no contribution from $\beta$-decay

Step 2: Apply atomic number conservation.
Initial atomic number:
\[ Z = 92 \]
Final atomic number from Pb:
\[ Z = 82 \]
Contribution from 7 $\alpha$ particles:
\[ 7 \times 2 = 14 \]
So after $\alpha$ decay:
\[ 92 - 14 = 78 \]

Step 3: Compare with final atomic number.
Final atomic number required is 82, but we have 78

Step 4: Effect of $\beta^-$ decay.
Each $\beta^-$ decay increases atomic number by 1
\[ 78 + m = 82 \]

Step 5: Solve for $m$.
\[ m = 4 \]

Step 6: Conclusion.
\[ \boxed{4} \]