Respuesta :
Answer:
(a) 8.3 g (b) 2.8 g (c) 0.0 g
Explanation:
We can express a decay model as
[tex]N(t)=N_0*e^{-\lambda t}[/tex]
(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
[tex]N(3.8)=\frac{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[/tex]
Now we have the model
[tex]N(t)=10*e^{-0.1824 t}[/tex]
(a) In one day the amount of radon will be 8.3 grams
[tex]N(1)=10*e^{-0.1824 *1}=10*0.833=8.3[/tex]
(b) In one week the amount of radon will be 2.8 grams
[tex]N(1)=10*e^{-0.1824 *7}=10*0.279=2.8[/tex]
(c) In one year the amount of radon will be 0.0 grams
[tex]N(1)=10*e^{-0.1824 *365}=10*(1.22*10^{-29})=1.22*10^{-28} \approx 0[/tex]
