# Quantum yield and spontaneous decay

I'm trying to figure out how many atoms are decaying spontaneously in a span of 2 seconds. Let's say that the quantum yield is 0.45, and that the lifetime "τ" (tau) is 10 microseconds.

Then I found that the radiative lifetime is 22.2 microseconds. However, at this point I'm stuck. I don't know a relationship in order to get the amount of spontaneously decayed atoms.

What relationship is there in order to get spontaneous decayed atoms with this information? Also, this is homework, so just a hint would be appreciated.

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You can use the equation

$$N_2(t) = N_2(0)\exp\biggl(-\frac{t}{\tau}\biggr)$$

if you can find a value for $N_2(0)$ which is the population at $t=0$.

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Hi Steven - please keep in mind that this site has a global audience. Anything personally specific to you and the original asker doesn't belong in an answer. I've edited your answer accordingly. –  David Z Feb 10 '13 at 23:01
@Steven: Yes, that's exactly what I thought too. Except that I couldn't find a way to find population at the initial time. –  Chris Harris Feb 10 '13 at 23:30
In this case I would guess that quantum yield is supposed to be the average number of photons created per decay. So the radiative lifetime is not 10$\mu$s/0.45, it's 10$\mu$s, but when you count the number of photons emitted in 2 seconds you need to divide this number by 0.45 to get the number of decays.
We need a bit more information to answer that. If the half life is 10$\mu$s then by the time 2s has elapsed virtually all the atoms will have decayed. Presumably in your system something is exciting atoms and they then decay. –  John Rennie Feb 10 '13 at 7:21