Question #e130f

1 Answer
Somebody N.
Dec 2, 2017

See below.

Explanation:

i).

You know the half life is 8 years, and after this time the substance will have halfed. From our equation using R as the initial amount and #1/2 R# as the amount after 8 years:

#R/2=Re^(-8k)#

Now we solve this, to reach the desired form:

Divide by R:

#1/2=e^(-8k)#

Taking natural logs of both sides:

#ln(1/2)=-8klnecolor(white)(888)# ( #lne=1#)

#ln(1/2)=-8k#

Divide by 8:

#1/8ln(1/2)=-k#
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ii).

We know from previous answer that #k=-1/8ln(1/2)#

#R(t)=Re^(-(-1/8ln(1/2)))#

#R(t)=Re^((1/8ln(1/2)))#

#R(t)=Re^(ln(1/2)^(1/8))color(white)(8888)# ( #e^ln(1/2)=1/2# )

#R(t)=R_(o)(1/2)^(1/8)#

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iii).
Since our original substance was 100% and 75% has decayed, we would expect to be left with 25% or #1/4# of the original amount. So, using our equation:

#R/4=Re^(1/8ln(1/2)t#

Solving for #t#:

#1/4=e^(1/8ln(1/2)t#

#1/4=(1/2)^(1/8t)#

Taking logs:

#ln(1/4)=1/8tln(1/2)color(white)(8888)#(from previous answer)

#8((ln(1/4))/(ln(1/2)))=t=>t=16#

16 years

This could have been worked out by using simple logic. If we lose 50% in 8 years, then in another 8 years we would lose 50% again.

50% of 50% is 25%.

I included the working anyway, because some teachers think you should go through all the steps, even though it's unnecessary.

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iv).

Amount after 20 years is 2 grams.

#:.#

#2=Re^(1/8ln(1/2)(20)#

#2=Re^(5/2ln(1/2)#

Dividing by R:

#2/R=e^(5/2ln(1/2)#

#2/R=e^(ln(1/2)^(5/2)#

#2/R=(1/2)^(5/2)=(sqrt(1))/(sqrt(32))=1/(4sqrt(2))=sqrt(2)/8#

#->2/R=sqrt(2)/8=>R=16/sqrt(2)~~11.31# grams ( 2 .d.p.)

Check:

#R(t)=(11.31)e^(5/2ln(1/2))=1.999344423#

Very close to 2 grams. We wouldn't expect it to be exact, since we rounded to 2 d.p.

Hope this helps you.

The question looks scary at first, but once you break it down it's not so difficult.

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