Physics Help Forum Ratio of N between two atoms

 Nuclear and Particle Physics Nuclear and Particle Physics Help Forum

 May 30th 2019, 03:18 AM #1 Junior Member   Join Date: May 2019 Posts: 4 Ratio of N between two atoms 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.
May 30th 2019, 03:28 AM   #2

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 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
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 May 30th 2019, 04:22 AM #3 Junior Member   Join Date: May 2019 Posts: 4 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.
 May 30th 2019, 05:46 AM #4 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 373 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: $\displaystyle \frac{A_0}{B_0} = 0.052 \frac{21.4}{14.4} \cdot \frac{\exp\left(-\frac{77.5}{14.4}\right)}{\exp\left(-\frac{77.5}{21.4}\right)} = 0.014$ topsquark and CBM like this. Last edited by benit13; May 30th 2019 at 05:51 AM.
May 30th 2019, 08:41 AM   #5

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 Originally Posted by benit13 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
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