Please review a rate of decay problem for me

In summary: Just seemed simpler to me.In summary, the half-life of Cobalt-60 is 5.24 years. A 100 mg sample would decay to 1 mg over 5.24 years.
  • #1
Permanence
53
2
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

3525291ec72b23583c072534123240cb.png
 
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  • #2
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
Yep. Thank you for your feedback. I'll email the TA tonight.
 
  • #4
Permanence said:
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

3525291ec72b23583c072534123240cb.png

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
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:

Related to Please review a rate of decay problem for me

1. How do you calculate rate of decay?

The rate of decay can be calculated by dividing the change in the quantity of a substance over a certain period of time by the initial quantity. This is represented by the equation: rate of decay = (final quantity - initial quantity) / initial quantity.

2. What is the formula for exponential decay?

The formula for exponential decay is: y = a(1 - r)^t, where y is the final quantity, a is the initial quantity, r is the rate of decay, and t is the time elapsed.

3. How does half-life relate to rate of decay?

Half-life is the amount of time it takes for half of the initial quantity of a substance to decay. The rate of decay is directly related to the half-life, as a higher rate of decay will result in a shorter half-life and vice versa.

4. Can rate of decay be negative?

No, rate of decay cannot be negative. It represents the decrease in quantity over time, so it will always be a positive value.

5. What factors can affect the rate of decay?

The rate of decay can be affected by factors such as temperature, pressure, and the presence of other substances. It can also vary depending on the type of substance and the environment it is in.

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