Question

Alexander Litvinenko was poisoned with 10 micrograms of the radioactive substance Polonium-210. Since radioactive decay follows a compounded continuously model, we can determine the amount of substance left in Alexander Litvinenko's body at any given time. If Polonium-210 has a decay rate of .502%, then determine the amount of Polonium-210 left in his body after 72 days. Provide 3 decimal places and a label in your answer.

Answer 1

The amount of Polonium-210 left in his body after 72 days is 6.937 μg.

Explanation:

The decay rate of Polonium-210 is the following:

`N(t) = N_(0)e^(-\lambda t)` (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:

`\lambda = (ln(2))/(t_(1/2))` (2)

Where t(1/2) is the half-life of Po-210 = 138.376 days

By entering equation (2) into (1) we have:

`N(t) = N_(0)e^{-(ln(2))/(t_(1/2))*t}} = 10* (99.498)/(100)*e^{-(ln(2))/(138.376)*72} = 6.937 \mu g`

Therefore, the amount of Polonium-210 left in his body after 72 days is 6.937 μg.

I hope it helps you!