Phosphorus-32 (P-32) has a half-life of 14.2 days. If 200 g of this substance are present initially, find the amount Q(t) present after t days. (Round your growth constant to four decimal places.)
How fast is the P-32 decaying when t = 21.5? (Round your answer to three decimal places.)
7 answers
Now I did problems very similar to this for you earlier. You try and I will check if I am around.
This is what I got and it says it is wrong. Can you tell me what I'm doing wrong please?
Thanks
Q(t) = 200(.5)^(.0704t)
Q '(t) = (200/14.2) (.5)^(t/14.2)
so when t = 21.5
Q ' (t) = (200/14.2) (.5)^(21.5/14.2)
= 4.931 g/day
Thanks
Q(t) = 200(.5)^(.0704t)
Q '(t) = (200/14.2) (.5)^(t/14.2)
so when t = 21.5
Q ' (t) = (200/14.2) (.5)^(21.5/14.2)
= 4.931 g/day
A = Ai e^-kt
.5 = e^-14.2 k
ln .5 = -14.2 k
k = .0488
so
Q = 200 e^-.0488 t
.5 = e^-14.2 k
ln .5 = -14.2 k
k = .0488
so
Q = 200 e^-.0488 t
dQ/dt = 200(-.0488) e^-.0488 t
I got t=70 and it still says it is wrong.
ok I got now. Thank you so much for your help, I appreciate it.
Atta go :)
You are welcome.
You are welcome.
0 answers