# Please review a rate of decay problem for me

1. Oct 23, 2013

### Permanence

Thank you for taking the time to read my thread. I do not agree with the TA's correction of my quiz and would appreciate if someone could give it a look before I approach him about it.
1. The problem statement, all variables and given/known data
Cobalt-60 has a half-life of 5.24 years. How long would it take for a 100 mg sample of Cobalt-60 to decay to 1 mg?

2. Relevant equations
y = sample*e^(-k*t)
I assumed that k was negative because it was a decay problem. My TA did not agree. I also assumed that the answer must be positive, and it appears that ln(1/100) is indeed negative. We did not have calculators for this quiz.

3. The attempt at a solution
k = -ln(2) / 5.24

y = 100 * e^-[(ln(2)/5.24)*t]
t = ln(1/100) / [ln(2)/5.24) ; where t is expressed in years

2. Oct 23, 2013

### haruspex

If the red lines converting minus signs to plus signs are the TA's correction, sack the TA. Since ln(1/100) is negative, a positive denominator will give a negative number of years. (And I agree with your answer.)

3. Oct 23, 2013

### Permanence

Yep. Thank you for your feedback. I'll email the TA tonight.

4. Oct 23, 2013

### Ray Vickson

You and the TA both made errors. If you write $y(t) = c e^{-kt}$ then $k > 0$ because you want y(t) to be a decreasing function. If you write $y(t) = c e^{kt}$ then you need $k < 0$ in order to have a decreasing function y(t).

You wrote it correctly in the hand-written work; you just made a mistake in your typed description above. However, the TA made something that is correct into something that is incorrect.

5. Oct 23, 2013

### Dick

I would have just started from 1=100*(1/2)^(t/5.24), using the direct definition of half-life and then let the logs sort themselves out. There's really no need to figure out k. Not to imply there's anything wrong with doing it the other way.

Last edited: Oct 23, 2013