Radioactive chain decay

[bitrex]

Homework Statement

I'm looking at a problem from MIT's Open Courseware on radioactive chain decay, i.e. one element decays into another decays into another, finding the quantity at time t.

Homework Equations

The standard linear differential equation governing exponential decay.

The Attempt at a Solution

I'd just like to make sure I'm going about this the right way - to calculate say the amount of substance 2 present at time T you'd solve the differential equation for the first substance finding $$N(t) = N_o {e}^-{\lambda t}$$, then take that and plug it back into the standard exponential differential equation as the quantity, solve that differential equation etc. God knows I'm not going to try to code it up in Latex, but I'd just like to know that I'm on the right track.

 

[HallsofIvy]

What are the differential equations you refer to?

 

[bitrex]

The differential equation governing exponential decay, i.e. $$\frac{dN}{dT} = -\lambda N$$. Solving this through separation of variables should give me the amount of the first substance remaining at time T, that is $$N(t) = N_o {e}^{-\lambda t}$$. If I want to find the amount of the next product of the chain decay at time T, I assume I would substitute the second equation for the amount of the first substance at time T back into the original differential equation as N and solve that to get $$N_2(t)$$, the amount of the second substance...

 

[bitrex]

Actually, I think I might have to substitute $$(N_o - N_o e^{-\lambda t})$$ since at time T = 0 there isn't any of the second substance yet produced to begin decaying!

 

[HallsofIvy]

The "second product" would be the result of the decay of the first product and that will have a different "[/itex]\lambda[/itex]".
In the formula you give,
$$\frac{dN}{dt}= -\lambda N$$,
N is the amount of the original substance at time t. The "product" N1 is
$$N_0- N_1(t)= N_0(1- e^{-\lambda t}$$
as you say.