The half-life of cesium-137 is 30 years. Suppose we have a 100-mg sample.

Question

9. The half-life of cesium-137 is 30 years. Suppose we have a 100-mg sample.

(a) Find the mass that remains after years.
(b) How much of the sample remains after 100 years?
(c) After how long will only 1 mg remain?

Answer 1

Ok this is an exponential growth/exponential decay problem. 

 

The formula for this is y = Ce^kt

 

Suppose we have a 100 gram sample of Cesium-137. The 100 Grams is our C in the formula.  For t =30 and y =50.  The reason for this is because the problem tells us that after 30 years half of this will be gone.  Half of 100 is 50.  Now we can set up the equation. 

 

50=100e^k(30)

 

You divide by 100 to get 1/2=e^30k

 

Then you take a ln of both sides.  This will get you the value of k.  Remember ln and e cancel each other out so you will get

 

ln(1/2)=ln(e^30k))

 

This in turn will solve for k since ln and e cancel each other out.  The k is a constant that can set up your problem to solve any of these types of problems. 

 

After you do this you will get

 

-.693=30k   You will divide by 30 to get k=-.0231

 

The final equation for this problem is y=100e^-.0231t

 

This formula will help you solve for part c when you plug in 1 for y by using the same rules as above and taking natural log of both sides.  Hope this helps!