Question

What mass of `""^(55) "Cr"` must be delivered in `"mg"` to a patient if the total drug transport time is `"12 hours"` and the minimum dosage needed is `"1.0 mg"`? The half-life is `"2 hours"`.

Answer 1

`6` half-lives pass, so you need the mass to be able to decay to half its previous amount 6 times before you reach `"1.0 mg"`.

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The idea is that the `""^55 "Cr"` is inconveniently turning into something unusable, and the usable quantity halves every `2` hours.

By definition, one half-life passing by gives:

`m_(""^55 "Cr") = 1/2m_(""^55 "Cr",0)`

where `m_(""^55 "Cr")` is the current mass and `m_(""^55 "Cr",0)` is the starting mass.

When `\mathbf(n)` number of half-lives pass by, however, we have:

`m_(""^55 "Cr") = (1/2)^nm_(""^55 "Cr",0)`

`= 1/(2^n)(m_(""^55 "Cr",0))`

Then, the total time required to transport is `t`, and each half-life that passes by takes `t_"1/2"` hours to pass. So, `n = t/(t_"1/2")`.

`color(green)(m_(""^55 "Cr") = 1/(2^(t"/"t_"1/2"))(m_(""^55 "Cr",0)))`

So, we have to ensure that `m_(""^55 "Cr")`, the mass delivered, is a little above `"1.0 mg"`, by starting with an initial mass `m_(""^55 "Cr",0)` that doesn't halve in quantity too many times.

`=> "1.0 mg" = 1/(2^(12 cancel"hours""/"2 cancel"hours"))(m_(""^55 "Cr",0))`

`"1.0 mg" = 1/(64)(m_(""^55 "Cr",0))`

`=> color(blue)(m_(""^55 "Cr",0) ~~ "64.0 mg")`