Question

What is the half-life of radon-222 if a sample initially contains 120.0 mg and only 18.1 mg after 12.3 days?

Answer 1

Answer:

$t_{1/2}=4.51days$

Explanation:

Hello.

In this case, since the kinetics of radioactive decay in terms of initial and final amounts, elapsed time and half-life is:

$A=A_0\times2^{-t/t_{1/2}}$

we can compute the half time as shown below:

$2^{-t/t_{1/2}}=\frac{A}{A_0}\\\\-\frac{t}{t_{1/2}} ln(2)=ln(\frac{A}{A_0})\\\\t_{1/2}=-\frac{t\times ln(2)}{ln(\frac{A}{A_0})}$

Plugging in the known amounts and elapsed time, we obtain:

$t_{1/2}=-\frac{12.3\times ln(2)}{ln(\frac{18.1}{120})}\\\\t_{1/2}=4.51days$

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