An exponential decay problem.

Printable View

May 4th 2010, 07:47 PM
kmjt

An exponential decay problem.

A sample of uranium-239 (U-239) decays into neptunium-239 (Np-239) according to the standard decay function:
http://img199.imageshack.us/img199/1849/expfunction.png
After 10 minutes, the sample has decayed to 64% of its initial amount.

a) The value of the disintegration constant, pi, is approximately 0.045/min

b) The half-life of U-239 is 15.5 minutes approximately.
(working on this at the moment)

c) Write the equation that gives the amount of U-239 remaining as a function of time, in terms of its half-life.

d) Suppose the initial sample had a mass of 25mg. How fast is the sample decaying after 15 minutes?
May 4th 2010, 08:18 PM
pickslides

Is this equation $N(t)= N_0e^{-\pi t}$

$N_0$ is the initial amount of uranium at $t= 0$

You are told that at $t = 10$ then $N(10) = 0.65\times N_0$

So you have

$N(t)= N_0e^{-\pi t}$

at $t = 10$

$0.65\times N_0= N_0e^{-10\pi }$

Dividing $N_0$ from each side leaving

$0.65= e^{-10\pi }$

Can you now solve for $\pi$ ?
May 5th 2010, 04:25 PM
kmjt

Solved for pi and got approximately 0.043. I believe that is correct.

For part b how would I start? (Worried)
May 5th 2010, 04:33 PM
skeeter

Quote:

Originally Posted by kmjt

Solved for pi and got approximately 0.043. I believe that is correct.

For part b how would I start? (Worried)

half-life ... time it takes for half of the original amount ( $N_0$ ) to decay.

$\frac{N_0}{2} = N_0 e^{-\pi t}$

solve for $t$
May 5th 2010, 04:43 PM
pickslides

Quote:

Originally Posted by kmjt

Solved for pi and got approximately 0.043. I believe that is correct.

For part b how would I start? (Worried)

For $N(t)= N_0e^{-0.043 t}$ solve for $t$ when $N(t) = \frac{N_0}{2}$

I.e $\frac{N_0}{2}= N_0e^{-0.043 t}$

Go get em!
May 5th 2010, 04:45 PM
kmjt

Hmm, this is how I tried solving for t:

As you said: No/2 = Noe^-pi(t)

I rewrote it as (0.5)( No) = Noe^-pi(t)

Divided No out..

0.5 = e^-pi(t)

ln(0.5) = ln(e^-pi(t))

ln(0.5) = -pi(t)

ln(0.5)/-pi = t

t = approximately 0.22

Where did I go wrong? The answer is 15.5 minutes (Thinking)

Just subbed in 0.043 for pi and the answer I got was -7..
May 5th 2010, 05:02 PM
skeeter

Quote:

Originally Posted by kmjt

Hmm, this is how I tried solving for t:

As you said: No/2 = Noe^-pi(t)

I rewrote it as (0.5)( No) = Noe^-pi(t)

Divided No out..

0.5 = e^-pi(t)

ln(0.5) = ln(e^-pi(t))

ln(0.5) = -pi(t)

ln(0.5)/-pi = t

t = approximately 0.22

Where did I go wrong? The answer is 15.5 minutes (Thinking)

Just subbed in 0.043 for pi and the answer I got was -7..

you used the constant $\pi = 3.14159...$

also, a correction ... $\pi = \frac{\ln(.64)}{-10} \approx 0.0446...$
May 5th 2010, 06:17 PM
kmjt

Ooooh, worked it out and got the right answer (Clapping)

The next part asks for an equation that gives the amount of U-239 remaining as a function of time, in terms of its half-life.

The answer is N( t) = N0(1/2)^(t/15.4)

What are the steps one must take to get to this answer? And i'm looking at the exponent part.. t/15.4, and i'm wondering why is it not 15.5? Or is it something completely different?
June 5th 2010, 01:10 PM
kmjt

Quote:

Write the equation that gives the amount of U-239 remaining as a function of time, in terms of its half life.

The answer to c) is:

http://img268.imageshack.us/img268/6...ionequatio.png

But what steps would I have to go through to get to that answer?