write an exponential function for a decay problem

Question

Please let me know if this expoinential function is correct for this story problem,  y = 100 (1 -.5)^x

 

Andrew measures the amount of a very unstable substance to be 100 moles.  The half-life of this substance is 3 days (after 3 days, half is gone).

 

Write an exponential function that models this situation where, y, is the amount of substance and x is time in days.

 

How would I calculate the equation to complete this charrt:

 

 

                      Days               Amount

                 _______________________

 

               -3 (3 days prior)         ??       (I think this would be 200, but I don't understand how to calculate it.

 

                     0                         ??

 

                     3                         ??

 

                     6                         ??

 

 

Thanks!

Answer 1

In general, whenever radioactive decay is being measured

 

A_(t) = A₀e^-kt where A0 is the initial amount (in this case 100 moles), A_(t) is the amount remaining after time t has elapsed, t is the elapsed time, and k is a constant (yet to be determined).

 

We know that the radioactive material has a half-life of 3 days.  We can use that fact to calculate k.

 

When 3 days have elapsed, the original 100 moles has been reduced to only 50 moles.  In other words

 

e^-kt = (1/2)

 

Replacing t with 3 days gives

 

e^-3k = (1/2)

 

-3k = ln(1/2)

 

from which

 

k = 0.231 days^-1

 

Let's complete the table:

 

Days          Amount

 

  -3             199.971 moles    [A_(-3) = A₀e-kt = (100 moles)e-^{(-3 days)(0.231)} = 199.971 moles]

 

    0             100 moles           [A₀ = A₀e^-kt = (100 moles)e^(0)*(0.231) = 100 moles

 

    3                50 moles           [A_{(3 days)} = A₀e^-kt = (100 moles)e^{-(3 days)*(0.231)} = 50 moles]

 

    6                25 moles           [A_{(6 days)} = A₀e^-kt= (100 moles)e^{-(6 days)*(0.231)} = 25 moles]

 

 

 

Answer 2

Your equation is pretty close to the right one.  you never accounted for the 3 days it takes to become 1/2

 

The equation would be y = 100(.5)^x/3

 

To fill the table you would plug in the time value for x and calculate it on your own or with a calculator.  for example

 

for x = -3

y = 100(.5)^-3/3

y = 100(.5)^-1

y = 100(2)

y = 200 moles

 

 

for x = 0

y = 100(.5)^0/3
y = 100(.5)⁰
y = 100(1)
y = 100 moles

 

Why don't you try the other values?