Cobalt-60 decay problem: is k negative?

[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.

Homework Statement

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?

Homework 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.

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

 

[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.)

 

[Permanence]

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

 

[Ray Vickson]

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.

Homework Statement

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?

Homework 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.

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

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.

 

[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.