A research assistant made 160 mg of radioactive sodium (Na^24) and found that there was only 20 mg left 45 h later, how much of the original 20 mg would be left in 12 h?

2 Answers
Aug 6, 2016

#=11.49# mg will be left

Explanation:

Let rate of decay be #x# per hour
So we can write
#160(x)^45=20#
or
#x^45=20/160#
or
#x^45=1/8#
or
#x=root45(1/8)#
or
#x=0.955#
Similarly after #12# hours
#20(0.955)^12#
#=20(0.57)#
#=11.49# mg will be left

Aug 6, 2016

Just to use the conventional radioactive decay model as a slight alternative method.

After 12hr we have 11.49mg

Explanation:

Let #Q(t)# denote the amount of sodium present at time #t#. At #t=0, Q = Q_0#

It's a fairly simple model to solve with ODEs but as it's not really related to the question, we end up with

#Q(t) = Q_0e^(-kt)# where #k# is a rate constant.

First we find the value of #k#

#Q_0 = 160mg, Q(45) = 20mg#

#Q(45) = 20 = 160e^(-45k)#

#therefore 1/8 = e^(-45k)#

Take natural logs of both sides:

#ln(1/8) = -ln(8) = -45k#

#k=(ln(8))/45 hr^(-1)#

#therefore Q(t) = Q_0e^(-(ln(8))/45t)#

So starting with #Q_0 = 20mg#

#Q(12) = 20e^(-(ln(8))/45*12) = 11.49mg#