Question

Radon gas that may be found in US buildings/homes has a half-life of about 3.8 days. If a 2. residential home had 10 grams radon, how much will be present after living in the house for a. One day b. One week c. One year

Answer 1

(a) 8.3 g (b) 2.8 g (c) 0.0 g

Step-by-step explanation:

We can express a decay model as

`N(t)=N_0*e^(-\lambda t)`

(N(t) is the grams of Radon in time t, N0 is the initial grams at t=0, lambda is the decay parameter, that we have to calculate.

With the half life period, we know that in 3.8 days we will have half the initial amount of Radon

`N(3.8)=(N_0)/(2)=N_0*e^(-\lambda*3.8)  \\\\(1/2)=e^(-\lambda*3.8) \\\\ln(1/2)=-\lambda*3.8\\\\\lambda=-ln(1/2)/3.8=0.693/3.8=0.1824`

Now we have the model

`N(t)=10*e^(-0.1824 t)`

(a) In one day the amount of radon will be 8.3 grams

`N(1)=10*e^(-0.1824 *1)=10*0.833=8.3`

(b) In one week the amount of radon will be 2.8 grams

`N(1)=10*e^(-0.1824 *7)=10*0.279=2.8`

(c) In one year the amount of radon will be 0.0 grams

`N(1)=10*e^(-0.1824 *365)=10*(1.22*10^(-29))=1.22*10^(-28) \approx 0`