What Are the Decay Dynamics of a Radioactive Substance Over Time?

[FinSanity]

[SOLVED] A radoactive substance

Homework Statement

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?

Homework Equations

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

c) A=A0e^-λt

b) A=λN => N = A/λ

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

 

[dynamicsolo]

Homework Statement

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?

Homework Equations

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

c) A=A0e^-?t

b) A=?N => N = A/?

The Attempt at a Solution

a) -? = ((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

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.