Find the amount of `""_(92)U^(238)` required to produce a-particles of 20 millicurie strength. The half-life of `""_(92)U^(238)` is `4.5 xx 10^(9)` years.

Explanation

To find the amount of Uranium-238 (`""_(92)U^(238)`) required to produce alpha particles of 20 millicurie strength, we can follow these steps:

### Step 1: Convert Millicurie to Disintegrations per Second
1. **Conversion Factor**: 1 Curie (Ci) = \(3.7 \times 10^{10}\) disintegrations per second (dps).
2. **Calculate Activity in dps**:
\[ \text{Activity (A)} = 20 \, \text{mCi} = 20 \times 10^{-3} \, \text{Ci} = 20 \times 10^{-3} \times 3.7 \times 10^{10} \, \text{dps} \]
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