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

`t_(1/2)=4.51days`

Step-by-step 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*2^{-t/t_(1/2)}`

we can compute the half time as shown below:

`2^{-t/t_(1/2)}=(A)/(A_0)\\\\-(t)/(t_(1/2)) ln(2)=ln((A)/(A_0))\\\\t_(1/2)=-(t* ln(2))/(ln((A)/(A_0)))`

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

`t_(1/2)=-(12.3* ln(2))/(ln((18.1)/(120)))\\\\t_(1/2)=4.51days`

Best regards!