# Homework Help: Radioactive decay equilibrium when decay constants are equal

1. Sep 17, 2006

### siifuthun

I need a refresher on calculus need to derive this equation, the question is:

A --> B --> C
and where decay constants:
http://img183.imageshack.us/img183/4403/aki2.jpg [Broken]
Derive an expression for the activity of B as a function of time.

So what I'm having trouble with is changing the derivation for the number of N when lambda_A = lambda_B so that I derive the activity of B.
Normally we would have:
http://img157.imageshack.us/img157/3501/ajh6.jpg [Broken]
But because of the condition, we have to go about it differently. So setting the decay constants equal to each other:
http://img182.imageshack.us/img182/192/asa5.jpg [Broken]

And this is where I get a little stumped trying to change the derivation. When lambda_A doesn't equal lambda_B, the derivation ends up with:
http://img226.imageshack.us/img226/1602/akk1.jpg [Broken]

The problem I'm having is that I'm not sure what has to be done differently in the original derivation to solve for activity, my calculus skills are a little rusty so if anyone has any suggestions or advice, it's greatly appreciated.

Last edited by a moderator: May 2, 2017
2. Sep 18, 2006

### Astronuc

Staff Emeritus
$$dN_B\,=\,\lambda_A\,(N^0_A\,e^{-\lambda_At}\,-\,N_B)\,dt$$

$$\frac{dN_B}{(N^0_A\,e^{-\lambda_At}\,-\,N_B)}\,=\,\lambda_A\,dt$$

which seems problematic.

But the expression on the left can be written

$$\frac{e^{\lambda_At}dN_B}{(N^0_A\,-\,e^{\lambda_At}N_B)}\,=\,\lambda_A\,dt$$

On the other hand, one does not find two radionuclides in a successive decay scheme having equal decay constants.

3. Sep 18, 2006

### Gokul43201

Staff Emeritus
There's two approaches to figuring this out (the easy way and the harder way).

The harder way is to look for solutions to the differential equation you've finally arrived at.

The easy way is to find the limit of the solution given for the general case, as $\lambda _B \rightarrow \lambda _A$

Hint: Write $\lambda _B = \lambda _A + \delta$ and find the limit of the first term of the general result (the one for unequal lambdas) as $\delta \rightarrow 0$. Use the Taylor expansion for the small exponential, and the limit pops out quite happily.