What will be the activity of 0.5 g sample of `""_(92)U^(238)` if its half-life against a-decay is `1.5 xx 10^(17)` s?

Explanation

To find the activity of a 0.5 g sample of Uranium-238 (U-238) with a half-life of \(1.5 \times 10^{17}\) seconds, we can follow these steps:

### Step 1: Calculate the decay constant (\(\lambda\))
The decay constant is related to the half-life (\(t_{1/2}\)) by the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Given that \(t_{1/2} = 1.5 \times 10^{17}\) s, we can substitute this value into the equation. The natural logarithm of 2 (\(\ln(2)\)) is approximately 0.693.
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