The half-life of a radioactive kind of lead is 27 minutes. If you start with 88 grams of it, how much will be left after 54 minutes?

topic: Exponential growth and decay: word problems

1 Answer
Somebody N.
Mar 31, 2018

#22"grams"#

Explanation:

To solve this we need to find an equation in the form:

#(A)t=(A_0)e^(kt)#

Where:

#(A)t# is the amount after time #t#

#(A_0)# is the amount at the beginning at #t=0#

#k# is the growth/decay factor.

#t# is the time in this case minutes.

To find some of these values we use the given information.

If the half life is 27 minutes and we start with 88 grams then after 27 minutes we would expect to have 44 grams remaining.

So:

#(A_0)=88#

#(A)t=44#

#t=27#

Using these values:

#44=88e^(27k)#

We now solve for #k#:

Divide by 88:

#44/88=e^(27k)#

#1/2=e^(27k)#

Taking natural logarithms of both sides:

#ln(1/2)=27kln(e)#

#ln(e)=1color(white)(8888)# ( The logarithm of the base is always 1 )

#ln(1/2)=27k#

Divide by 27:

#ln(1/2)/27=k#

So we now have:

#At=88e^((ln(1/2)t)/27)#

This can be simplified using the laws of indices and logarithms:

#(A)t=88e^(ln(1/2)^(t/27)#

#(A)t=88(1/2)^(t/27)#

#(A)t=88(2)^(-t/27)#

Amount after 54 minutes:

#t=54#

#(A)t=88(2)^(-54/27)=22#

#22 \ \ \ "grams"# remaining after 54 minutes.

This could have been solved just by recognising that if the half life is 27 minutes then #54=2xx27-> 88/4=22#

i.e. after 27 minutes we have have the amount remaining in another 27 minutes we have half of the half or a quarter remaining.

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