Question

The radioactive decay is given as: \[ {}^{238}_{92}U \rightarrow {}^{210}_{82}Pb + 7\alpha + x\beta^{-}. \] Find the value of $x$.

Hint:

Remember: $\alpha$ emission decreases atomic number by 2. $\beta^{-}$ emission increases atomic number by 1. Always balance both mass number and atomic number.

Answer 1

Correct Answer: 4

Step 1: Understand the decay process.
Each $\alpha$-particle emission reduces:
Mass number by 4
Atomic number by 2
Each $\beta^{-}$ emission increases:
Atomic number by 1
Mass number remains unchanged
Step 2: Apply mass number conservation.
Initial mass number = 238.
After 7 $\alpha$ emissions:
\[ 238 - 7(4) = 238 - 28 = 210. \]
Final nucleus has mass number 210, which matches ${}^{210}Pb$.
Thus mass balance is satisfied.
Step 3: Apply atomic number conservation.
Initial atomic number = 92.
After 7 $\alpha$ emissions:
\[ 92 - 7(2) = 92 - 14 = 78. \]
Let number of $\beta^{-}$ emissions be $x$.
Each $\beta^{-}$ increases atomic number by 1.
Final atomic number becomes:
\[ 78 + x. \]
Given final atomic number of Pb = 82.
\[ 78 + x = 82. \]
\[ x = 4. \]
Final Answer:
\[ \boxed{x = 4}. \]