Question

If the activity of a radioactive substance is initially 398 disintegrations/minute and two days later it is 285 disintegrations/minute, what is the activity four days later still, or six days after the start? Give your answer in disintegrations/minute.

note: I'm really unsure of the step by step instructions on how to enter this in the calculator. Prof had an answer of N = 146 days with a disintegration constant of 0.16698/day. Please be very descriptive in your answer! *also what does the e represent in the formula N(t) = No e ^-ln2 x t/T1/2

Much appreciated! Thank-you!

Answer 1

ANSWER

Exercise Information:

A(0) = 398 disintegrations/minute

A(2) = 285 disintegrations/minute

A(4) = ?

Keep in mind that the activity of a radioactive substance is defined as the number of disintegrations per unit time. Since radioactive decay is random, the number of disintegrations per unit time is proportional to the number of radioactive atoms; that is:

So the activity is:

Then, the solution for N(t) is:

So if we multiply this by , we get:

Where A(t) is the activity at time t and A_o is the initial activity. We need to determine (apply natural logarithm both sides of equation):

Then, we can plug four days into our equation to get the activity at that time:

Finally, the activity four days later will be:

A = 146 disintegrations/minute

P.D.: If you do the exercise, pass by pass like me, you can notice that the disintegration constant given for your Prof is the that we calculate. And the "e" represent in the formula is the exponential function.

Regards!