Answer :

Final answer:

After 135 minutes, approximately 53.76 mg of lead-214 are left.

Explanation:

Lead-214 undergoes radioactive decay, meaning it decreases over time. The decay of lead-214 follows an exponential decay formula:

[tex]\[N(t) = N_0 * e^(-λt)\][/tex]

Where:

- [tex]\(N(t)\)[/tex] is the amount of lead-214 left after time \(t\),

- [tex]\(N_0\)[/tex] is the initial amount of lead-214 (800 mg in this case),

- [tex]\(λ\)[/tex] is the decay constant (a characteristic of the radioactive material), and

- [tex]\(t\)[/tex] is the time in minutes.

First, we need to find the decay constant λ. The half-life of lead-214 is about 26.8 minutes, which means [tex]\(N_0/2 = 800 mg/2 = 400 mg\)[/tex] of lead-214 will be left after one half-life. Using this information, we can rearrange the half-life equation to find \(λ\):

[tex]\[λ = \frac{-ln(0.5)}{26.8}\][/tex]

Now, we can use this λ value to calculate the amount of lead-214 left after 135 minutes:

[tex]\[N(135) = 800 * e^{\left(\frac{-ln(0.5)}{26.8}\right)*135}\][/tex]

After solving this equation, we find that approximately 53.76 mg of lead-214 are left after 135 minutes.

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