# 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".

Jun 23, 2016

$6$ half-lives pass, so you need the mass to be able to decay to half its previous amount 6 times before you reach $\text{1.0 mg}$.

The idea is that the $\text{^55 "Cr}$ is inconveniently turning into something unusable, and the usable quantity halves every $2$ hours.

By definition, one half-life passing by gives:

${m}_{\text{^55 "Cr") = 1/2m_(""^55 "Cr} , 0}$

where ${m}_{\text{^55 "Cr}}$ is the current mass and ${m}_{\text{^55 "Cr} , 0}$ is the starting mass.

When $\setminus m a t h b f \left(n\right)$ number of half-lives pass by, however, we have:

${m}_{\text{^55 "Cr") = (1/2)^nm_(""^55 "Cr} , 0}$

$= \frac{1}{{2}^{n}} \left({m}_{\text{^55 "Cr} , 0}\right)$

Then, the total time required to transport is $t$, and each half-life that passes by takes ${t}_{\text{1/2}}$ hours to pass. So, $n = \frac{t}{{t}_{\text{1/2}}}$.

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

So, we have to ensure that ${m}_{\text{^55 "Cr}}$, the mass delivered, is a little above $\text{1.0 mg}$, by starting with an initial mass ${m}_{\text{^55 "Cr} , 0}$ 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")