If the radon concentration has become 1/20th of what it started as, and its half-life is 3.8 days, what amount of time in days has passed?
1 Answer
Dec 24, 2017
The time is
Explanation:
The half life of Radon is
The radioactive constant is
The equation for radioactive decay is
#(N(t))/(N_0)=e^(-lambdat)#
Therefore,
#e^(-lambdat)=(N(t))/N_0=1/20# .
Taking natural logs of both sides,
#-lambdat=ln(1/20)=-ln20#
#t= (ln 20)/(lambda)#
#= ln20/(ln2//t_(1//2))#
#= ln20/ln2*t_(1//2)#
#= ln20/ln2*"3.8 days"#
#=# #"16.4 days"#