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?

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Answer 1 · 2017-12-24T10:28:50

The time is #"16.4 days"#

The half life of Radon is #t_(1//2)="3.8 days"#

The radioactive constant is #lambda=ln2/t_(1//2)#

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"#