# Half-life Stuff

1. Feb 22, 2006

### runicle

Something has a half-life of 18.4 days. How much time will a 50-mg sample take to decompose to 10 mg?
My steps
50(1/2)^n/18.4 = 10(1/2)^n/18.4
25^n/18.4 = 5^n/18.4
(5^2)^n/18.4 = 5^n/18.4
5^2n/18.4 = 5^n/18.4
Exponent view
2n/18.4 = n/18.4
2n-n = 18.4/18.4
n=1
Am i doing something wrong?

2. Feb 22, 2006

### stunner5000pt

half life equation is
$$q(t) = q_{0} 2^{-t/\tau}$$
where q(t) is the quantity after time t,
q0 is the initial quantity
tau is the half life

3. Feb 23, 2006

### konartist

Y=CoE^kt
Co = initial amount
Y= what amount is left
K= rate
T= time
First you have to find the rate
Since in 18.4 days you will have 1/2 left you set it up like this:

1/2Co=CoE^kt
Now the Co's cancel and you're left with :
1/2=e^kt
t=18.4
1/2=e^k(18.4)
Take the natural log of both sides now
ln(.5)=lne x 18.4k
-.6931471806 = 1 x 18.4k
divide by 18.4
k = -.0377
Now you must find the time it takes to go from 50 to 10.
10=50e^kt
k we know = -.0377 and we're looking for t

You should be able to figure it out from here.

(sorry if my work doesn't look fancy, I'm new to this stuff - I'll pick it up eventually)

4. Feb 23, 2006

### HallsofIvy

Staff Emeritus
This: 50(1/2)^n/18.4 = 10(1/2)^n/18.4 is your error.
You want to end with 10 g so you want
50(1/2)^(n/18.4)= 10

5. Feb 23, 2006

### runicle

Thank you HallsOfIvy for being specific for me:)