Answer :
Answer:
20.9 min time taken forn 99.0% of a sample of pure [tex]^{208}Tl[/tex].
Explanation:
From the given,
half line of [tex]^{208}Tl[/tex] = 3.1 min
[tex]t_{1/2}= 3.1min[/tex]
[tex]\lambda =\frac{ln(2)}{t_{1/2}}[/tex]
[tex]\lambda =\frac{ln(2)}{3.1} =0.22min^{-1}[/tex]
The ratio of the initial amount [tex]N_{o}[/tex] and amount in time is calculated as follows.
[tex]\frac{N}{N_{o}}=e^{-\lambda t}[/tex]
Now 99% got delayed, the left amount will be 1%.
[tex]\frac{N}{N_{o}}=\frac{1}{100}[/tex]
[tex]0.01=e^{-0.22min^{-1}.t}[/tex]
[tex]ln(0.01)=(-0.22min^{-1}.t)[/tex]
[tex]t=\frac{-ln(0.01)}{-0.22min^{-1}}=20.9min[/tex]
Therefore, 20.9 min time taken forn 99.0% of a sample of pure [tex]^{208}Tl[/tex].