Question

Radon, `._(86)^(222)Rn`, is a radioactive gas that can be trapped in the basement of homes, and its presence in high concnetrations is a known health hazard. Radon has a half-life of `3.83 days`. A gas sample contains `4.0 xx 10^(8)` radon atoms initially.
(a) How many atoms will remain after `12 days` have passed if no more radon leaks in? (b) What is the initial activity of the radon sample?

Answer 1

(a) A more precise answer is obtained by first finding the decay cosntant from equation
`lambda =(0.693)/(T_(1//2))=(0.693)/(3.83 days)=0.181 days^(-1)` Taking `N_(0) =4 xx10^(8) `and the value of `lambda` just found to obtain the number N remaining after 12 days as
`N=N_(0)e^(-lambdat)=-(4.0xx10^(8) "atoms") e^(-(0.181 days-1)(12 daus))`
`=4.6 xx10^(7)"atoms"` This is vey close to our original estimate of `5.0 xx10^(7) "atoms"`.
(b) First, we must express the decay constant in units of `s^(-1)` Using this value of `lambda`, we find that the initial activity as
`R= lambda N_(0)=(2.09 xx10^(-6) s^(-1))(4.0 xx 10^(8))`
`=840 decays s^(-1) =840 Bq`.