1. Nov 14, 2007

### FinSanity

1. The problem statement, all variables and given/known data

A radioactive substance has an activity of 10mCi. After 4 hours the activity is 6mCi.
a) What is the decay constant and the half life?
b) How many atoms are there in the beginning?
c) What is the activity after 30 hours?

2. Relevant equations

a) A=A0e^-λt => -λ = ((ln A/A0)/t
T½=((ln 2)/λ

c) A=A0e^-λt

b) A=λN => N = A/λ

3. The attempt at a solution

a) -λ = ((ln 6/10)/4) = 0,13
T½ = ((ln 2)/0,13) = 5,33h

c) A=10*e^-0,13*30 = 0,2mCi

b) N = 10/0,13=0,77

2. Nov 17, 2007

### dynamicsolo

I don't know if you're still looking at this thread. I agree with parts (a) and (b) [I get 0.22 mCi].

For part (c), first off, your decay constant is in units of inverse hours right now. To convert this usefully to the number of atoms involved, the activity needs to be related to seconds, since 1 Curie is 3.7·10^10 decay/sec (Becquerels). So the decay constant becomes

-(lambda) = 0.1277 (hr^-1) · (1 hr/3600 sec) = 3.547·10^-5 sec^-1 .

This gives

N = (10·10^-3 Ci)·(3.7·10^10 Bq)/(3.547·10^-5 sec^-1)

= 1.04·10^13 nuclei.