To find out how much Polonium-210 remains after a certain period, we can use the formula for exponential decay based on the concept of half-life. The formula is given by:
\[ N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
where:
- \( N \) is the remaining quantity of the substance.
- \( N_0 \) is the initial quantity of the substance.
- \( t \) is the elapsed time.
- \( t_{1/2} \) is the half-life of the substance.
In this case:
- \( N_0 = 128 \) grams (the initial amount of Polonium-210).
- \( t_{1/2} = 138 \) days (the half-life of Polonium-210).
- \( t = 414 \) days (the time elapsed).
First, we need to determine how many half-lives have passed in 414 days. We can find this by dividing the total time by the half-life:
\[ \text{Number of half-lives} = \frac{t}{t_{1/2}} = \frac{414}{138} \approx 3 \]
This means that approximately 3 half-lives have passed.
Next, we can compute the remaining amount of Polonium-210 using the half-life formula:
\[ N = 128 \left( \frac{1}{2} \right)^3 \]
Calculating \( \left( \frac{1}{2} \right)^3 \):
\[ \left( \frac{1}{2} \right)^3 = \frac{1}{8} \]
Now we can substitute this back into the equation:
\[ N = 128 \times \frac{1}{8} = 128 \div 8 = 16 \]
Thus, after 414 days, approximately 16 grams of Polonium-210 would be left.
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