Ratio of N between two atoms

CBM

May 2019
4
0
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.
 

topsquark

Forum Staff
Apr 2008
3,055
651
On the dance floor, baby!
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
 

CBM

May 2019
4
0
Okay so the derivatives are DA/dt=-TAtAoe^-TAt. I think. then substitute in my values for t and T. Which leaves me with -1.86775*10^-15Ao=-1.116*10^-15Bo*0.052.
At t=0 doesn't A0 and Bo just equal the amount?
so -1.86775*10^-15/-5.8032*10^-17 = the ratio.

But this isn't correct and Idk where exactly I went wrong.
 
Oct 2017
661
332
Glasgow
Just a little typo in Topsquark's post. The formulae should be:

\(\displaystyle A = A_0 \exp\left(-\frac{t}{\tau_A}\right)\)
\(\displaystyle B = B_0 \exp\left(-\frac{t}{\tau_B}\right)\)

The lifetime is on the denominator. For \(\displaystyle t = \tau\), the population decreases by a factor \(\displaystyle 1/e\).

Constraint:

\(\displaystyle \frac{dA}{dt} = 0.052 \frac{dB}{dt}\)

at t= \(\displaystyle 7.75 \times 10^{-8}\) s

Derivatives are:

\(\displaystyle \frac{dA}{dt} = -\frac{A_0}{\tau_A} \exp\left(-\frac{t}{\tau_A}\right)\)
\(\displaystyle \frac{dB}{dt} = -\frac{B_0}{\tau_B} \exp\left(-\frac{t}{\tau_B}\right)\)

Substituting into constraint and rearranging for ratio \(\displaystyle \frac{A_0}{B_0}\):

\(\displaystyle \frac{A_0}{B_0} = 0.052 \frac{\tau_A}{\tau_B} \cdot \frac{\exp\left(-\frac{t}{\tau_B}\right)}{\exp\left(-\frac{t}{\tau_A}\right)}\)

All time quantities are ratios, so we can just specify them in nanoseconds when substituting:

\(\displaystyle \frac{A_0}{B_0} = 0.052 \frac{21.4}{14.4} \cdot \frac{\exp\left(-\frac{77.5}{14.4}\right)}{\exp\left(-\frac{77.5}{21.4}\right)} = 0.014\)
 
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topsquark

Forum Staff
Apr 2008
3,055
651
On the dance floor, baby!
Just a little typo in Topsquark's post. The formulae should be:

\(\displaystyle A = A_0 \exp\left(-\frac{t}{\tau_A}\right)\)
\(\displaystyle B = B_0 \exp\left(-\frac{t}{\tau_B}\right)\)
I knew something looked off. I blame it on the benadryl. (I gotta blame it on something!)

Thanks for the catch.

-Dan