SOLUTION: Use a calculator to help solve each problem. This regarding radioactive decay. Can anyone give me a hand? Is the formula the same as carbon-14 dating? In 2 years, 20% of radi

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Question 62494: Use a calculator to help solve each problem.
This regarding radioactive decay. Can anyone give me a hand? Is the formula the same as carbon-14 dating?
In 2 years, 20% of radioactive element
decays. Find its half life.
Thanks for your help!

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Use a calculator to help solve each problem.
This regarding radioactive decay. 
Can anyone give me a hand? Is the formula 
the same as carbon-14 dating? 
In 2 years, 20% of radioactive element
decays. Find its half life. 
Thanks for your help!

The problem can be stated equivalently this way:

After 2 years, 80% of a quantity of radioactive 
element remains.  After how many years will only 
50% of it remain? 

The formula for all exponential growth or decay is

A = Pert

It is a growth when r is positive and a decay when 
r is negative. P represents the original amount 
and A represents the final amount after t units of 
time.

Suppose we begin with P units of this radioactive 
element.  Then when t = 2 years, 20% of P decays, 
leaving 80% of it, or .8P units remain.

So we substitute A = .8P, and t = 2.

.8P = Per(2)

Divide both sides by P and write r(2) as 2r

.8 = e2r

Use the rule that says any equation of the
form A = eB can be rewritten B = ln(A). 
We may rewrite the above equation as

2r = ln(.8)

r = ln(.8)/2 = -.1115717757

Since this is a decay, we expected r to be 
negative.  Now we substitute this value of r
in the equartion

A = Pert

A = Pe-.1115717757t

Now we wish to know its half life, or how
many years it will take P units of the 
radioactive subatance to decay to only 50% 
of P units or .5P units.

Su we substitute .5P for A:

.5P = Pe-.1115717757t

Divide both sides by P

.5 = e-.1115717757t

Rewrite this equation as

-.1115717757t = ln(.5)

Divide both sides by -.1115717757

            t = ln(.5)/(-.1115717757)

            t = 6.212567439 years

So its half life is about 6.2 years.

Edwin