Answer:
The amount of Polonium-210 left in his body after 72 days is 6.937 μg.
Step-by-step explanation:
The decay rate of Polonium-210 is the following:
[tex] N(t) = N_{0}e^{-\lambda t} [/tex] (1)
Where:
N(t) is the quantity of Po-210 at time t =?
N₀ is the initial quantity of Po-210 = 10 μg
λ is the decay constant
t is the time = 72 d
The decay rate is 0.502%, hence the quantity that still remains in Alexander is 99.498%.
First, we need to find the decay constant:
[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex] (2)
Where t(1/2) is the half-life of Po-210 = 138.376 days
By entering equation (2) into (1) we have:
[tex] N(t) = N_{0}e^{-\frac{ln(2)}{t_{1/2}}*t}} = 10* \frac{99.498}{100}*e^{-\frac{ln(2)}{138.376}*72} = 6.937 \mu g [/tex]
Therefore, the amount of Polonium-210 left in his body after 72 days is 6.937 μg.
I hope it helps you!
