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"#.

1 Answer
Truong-Son N.
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 #"1.0 mg"#.

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")#

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