# How long will it take for radioisotope Technetium 99 to decrease to 0.0001%?

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You need to solve for time the following decay equation, such that:

`e^(-lambda*t) = A/(A_0)`

Since the problem provides `A/(A_0) = 0.0001% ` yields:

`e^(-lambda*t) = 0.000001`

The isotope `99Tc` has a specific half life of `t_(1/2) = 2.1*10^5 ` years, hence, you may find `lambda` , using the following equation, such that:

`t_(1/2) = (ln 2)/lambda => lambda = (ln 2)/(t_(1/2))`

`lambda = (ln 2)/(2.1*10^5) => lambda = 0.33*10^(-5)`

Replacing `0.33*10^(-5)` for `lambda` you may evaluate t, such that:

`e^(-0.33*10^(-5)*t) = 0.000001`

`ln (e^(-0.33*10^(-5)*t)) = ln 0.000001`

`-0.33*10^(-5)*t*ln e = ln 0.000001 => -0.33*10^(-5)*t = ln 0.000001`

`t = ( ln 0.000001)/(-0.33*10^(-5)) => t = 41.86*10^5` years

Hence, evaluating the time taken by `99Tc` isotope to decrease `0.0001%` , yields `t = 41.86*10^5` years.