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I'm writing up a lab report and have a question about the following formula

$$N = N_0e^{-\lambda t}$$

$N$ indicates the number of nuclei left after a time $t$ and $N_0$ indicates how much there was to begin with. In the experiment, we used a scintillation counter (consists of a crystal and a photomultiplier tube) to detect the gamma rays beginning emitted by a decaying metastable Ba-137 to Ba-137. After recording the "counts" $C$, we corrected for the background radiation counts $BG$ by calculating $C -BG$. Then we plotted $\ln(C-BG)$ vs $t$. We did this because from the above formula

$$C-BG = (C-BG)_0e^{-\lambda t}$$

Then,

$$\ln(C-BG) = \ln(C-BG)_0 - \lambda t$$

Thus a plot of $\ln(C-BG)$ vs $t$ yields a straight line with slope $-\lambda$. My question is, how does counts relate to $N$? I thought first formula above indicates the number of nuclei left. However, the detector is detecting gamma rays from decaying nuclei, i.e. counting how many nuclei are decay?

So why can I replace $N$ with counts?

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    $\begingroup$ The count rate $C$ is directly proportional to the number of emitters $N$ remaining in the sample. $\endgroup$
    – rob ♦
    Commented Apr 27, 2014 at 22:44
  • $\begingroup$ Note also that $N = N_0 e^{-\lambda t}$ is all decays (at all angles), the decays seen by your device will actually be $N = \ell \epsilon A (N_0 e^{-\lambda t})$ where $\ell$ is live time (very nearly 1.0), $\epsilon$ is the quantum efficiency of your detector (probably also near 1), and $A$ is acceptance (a number expressing what fraction of decay gammas actually hit the detector). But all those corrections are effectively constant, so they don't affect your measured lifetime. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Apr 27, 2014 at 23:17
  • $\begingroup$ Is your measurement $C$ the total number of counts accumulated since the beginning of the experiment, or is it the number of counts in a short time interval (like counts per second, for example)? $\endgroup$
    – David Z
    Commented Apr 28, 2014 at 0:25
  • $\begingroup$ @DavidZ $C$ is the number of counts in a short time interval. We made like 900 different recordings. Each recording was like 4 seconds. $\endgroup$
    – DWade64
    Commented Apr 28, 2014 at 5:18
  • $\begingroup$ @DWade64 I'd suggest editing the question to reflect that, as it's important information. $\endgroup$
    – David Z
    Commented Apr 28, 2014 at 18:26

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When you measure the number of decays in a short time interval, $C$, you're effectively measuring the rate $R$ at which decays happen, because $C - BG \approx R\Delta t$ for some fixed interval $\Delta t$. But the rate of decay is related to the number of nuclei remaining as follows:

$$R = -\frac{\mathrm{d}N}{\mathrm{d}t} = -\frac{\mathrm{d}}{\mathrm{d}t}N_0 e^{-\lambda t} = N_0\lambda e^{-\lambda t} = \lambda N$$

So since $R = \lambda N$, you can multiply both sides of $N = N_0e^{-\lambda t}$ by $\lambda$ and find that $R = R_0e^{-\lambda t}$. Then if you multiply both sides of that by your sampling time $\Delta t$, you get

$$C - BG \approx (C - BG)_0 e^{-\lambda t}$$

which justifies why you can plot $\ln(C - BG)$ versus $t$ and get a straight line.

Essentially, you're not really replacing $N$ with $C - BG$, but rather converting the equation for $N$ into an equation for $C - BG$ that happens to take the same form.

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