The half-life of sodium-24 is 14.9h. A medical facility buys a 50mg sample. How do you determine the exponential function that models the amount, A(t), of sodium-24 that will be present in the sample after t hours?

2 Answers
Somebody N.
Mar 1, 2018

#color(blue)(A(t)=(50)2^(-t/14.9))#

Explanation:

We need to find an equation of the form:

#A(t)=ae^(kt)#

Where:

#A(t)=# the amount after time #t# ( in milligrams )

#a=# the initial amount.

#k=# the growth/decay factor.

#t=# time ( in this case hours )

From the given information. If the half life is #14.9color(white)(8)#hrs, then we would expect the mass of sodium to have halved after this period.

Plugging this into our equation:

#25=50e^(14.9k)#

Dividing both sides by #50#

#25/50=50/50e^(14.9k)#

#1/2=e^(14.9k)#

Taking natural logs of both sides:

#ln(1/2)=14.9kln(e)#

#ln(e)=1color(white)(888)# ( the logarithm of the base is always 1 )

#ln(1/2)=14.9k#

By the laws of logarithms:

#ln(a/b)=ln(a)-ln(b)#

So:

#ln(1/2)=ln(1)-ln(2)=0-ln(2)#

#-ln(2)=14.9k#

#k=-ln(2)/14.9#

#A(t)=50e^((-ln(2)/14.9)t)#

Exploiting the laws of indices:

#A(t)=50e^(((ln(2))t)(-1/14.9))#

#A(t)=(50)2^(-1/14.9t)#

#color(blue)(A(t)=(50)2^(-t/14.9))#

Check:

#25"mg"# after 14.9 hours.

#A(t)=(50)2^(-(14.9)/14.9)#

#A(t)=(50)2^-1=25"mg"color(white)(88)# as expected.

Note:

If you leave this with #bbe# as a base, the results will not be exact. This is because #bbe# is an irrational number.

Narad T.
Mar 1, 2018

The function is #A(t)=50e^(-0.0465t)#

Explanation:

The equation for the radioactive decay is

#A(t)=A_0e^(-lambdat)#

#lambda# is the radioactive constant

#lambda=(ln2)/t_(1/2)#

The half life of #"Sodium-24"# is #t_(1/2)=14.9h#

The initial mass is #A_0=50mg#

Therefore,

#A(t)=50e^(-(ln2)/14.9)t#

#A(t)=50e^(-0.0465t)#

graph{50e^(-0.0465x) [-11.27, 61.8, -0.3, 36.22]}

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