Respuesta :
Answer:
Thus the time taken is calculated as 387.69 years
Solution:
As per the question:
Half life of [tex]^{3890}Sr\, t_{\frac{1}{2}}[/tex] = 28.5 yrs
Now,
To calculate the time, t in which the 99.99% of the release in the reactor:
By using the formula:
[tex]\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}[/tex]
where
N = No. of nuclei left after time t
[tex]N_{o}[/tex] = No. of nuclei initially started with
[tex]\frac{N}{N_{o}} = 1\times 10^{- 4}[/tex]
(Since, 100% - 99.99% = 0.01%)
Thus
[tex]1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}[/tex]
Taking log on both the sides:
[tex]- 4 = \frac{t}{28.5}log\frac{1}{2}[/tex]
[tex]t = \frac{-4\times 28.5}{log\frac{1}{2}}[/tex]
t = 387.69 yrs
