Radioactive Trace Decay Calculation
Apply first-order kinetics and logarithmic manipulation to find the remaining concentration.
Question:
Assuming $1 \, \mu\text{g}$ of trace radioactive element $X$ with a half-life of $30 \text{ years}$ is absorbed by a growing tree.
The amount of $X$ remaining in the tree after $100 \text{ years}$ is ________ $\times 10^{-1} \, \mu\text{g}$.
[Given: $\ln 10 = 2.303 \,;\, \log 2 = 0.30$]
Detailed Solution
Step 1: Write down the given values and formula
- Initial amount, $N_0 = 1 \, \mu\text{g}$
- Half-life, $t_{1/2} = 30 \text{ years}$
- Total time, $t = 100 \text{ years}$
- Find: $N_t$ (amount remaining)
Radioactive decay follows first-order kinetics. The integrated rate equation is:
Step 2: Equate the rate constant ($k$)
We also know that $k = \frac{0.693}{t_{1/2}} = \frac{2.303 \log 2}{t_{1/2}}$. Substitute this into our rate equation:
Notice how $2.303$ brilliantly cancels out from both sides!
Step 3: Solve for $N_t$
Now plug in the given value for $\log 2 = 0.30$ and solve:
Take the antilog of both sides (since it's base 10):
The question asks for the answer in the format $Y \times 10^{-1} \, \mu\text{g}$.
$0.1 = \mathbf{1} \times 10^{-1}$
The Exponent & Log Trick (Faster!)
You can bypass the large rate constant equation entirely by using the general half-life formula: $N_t = N_0 \times (1/2)^n$, where $n$ is the number of half-lives ($n = t / t_{1/2}$).
Now, just take the base-10 log on both sides:
If $\log(N_t) = -1$, then $N_t = 10^{-1} = 0.1$. Simple and clean!
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