Originally Posted by **CBM** lifetime of the 4p state of atom A is 21.4 ns and the lifetime of the 3s state of atom B is 14.4 ns.
At t = 77.5 ns the rate at which a sample of atom A in the 4p state is decaying is 5.20% of the rate at which a sample of atom B in the 3s state is decaying.
What's the ratio of the amounts of atom A in the 4p state and atom B in the 3s state at t = 0 s?
I have absolutely no idea how to start this, so just a shove in the right direction would be appreciated. |

Say that we have an amount of A at t = 0 s, $\displaystyle A_0$ and an amount of B at t = 0 s, $\displaystyle B_0$. Let's say that the half-life of A is $\displaystyle \tau _A$ and that of B is $\displaystyle \tau _B$.

At a given time t we know that $\displaystyle A = A_0 e^{- \tau_A ~t}$ and $\displaystyle B = B_0 e^{- \tau _B ~ t}$. And finally, at t = 77.5 s $\displaystyle \dfrac{dA}{dt} = 0.0520 \dfrac{dB}{dt}$

So, what are the derivatives? How can you relate these to the amounts of A and B?

See if you can fill in the details. If you need more help, just ask.

-Dan