= Radiochemical neutrino counting model
{title2=$\mu_s=\epsilon\eta\frac R\lambda(1-e^{-\lambda T})e^{-\lambda d}(1-e^{-\lambda W})$}
An idealized <radiochemical neutrino counting model> separates production, survival, recovery and readout. Suppose a freshly cleared target produces atoms at a constant rate $R$ during exposure $T$. Each atom has <nuclear decay constant> $\lambda=\log 2/h$, where $h$ is its <half-life>. The expected surviving population obeys
$$
\frac{dN}{dt}=R-\lambda N,\qquad N(0)=0,\qquad N(T)=\frac R\lambda(1-e^{-\lambda T}).
$$
The proof is to multiply the <differential equation> by $e^{\lambda t}$ and apply <integration>. The expression also follows by summing the <exponential decay> survival probabilities of atoms produced at every earlier time. Production does not imply eventual recovery: atoms can decay during exposure.
Let $\eta$ be the fraction recovered instantaneously at extraction, $d$ the subsequent delay before counting, $\epsilon$ the probability of recording a decay, and $W$ the counting duration. The expected signal pulses are
$$
\mu_s=\epsilon\eta N(T)e^{-\lambda d}(1-e^{-\lambda W}).
$$
With constant background rate $b$, the expected total is $\mu=\mu_s+bW$. During counting, the expected instantaneous pulse rate is $s(t)+b$, with $s(t)=\epsilon\eta\lambda N(T)e^{-\lambda(d+t)}$. Equal time windows therefore lose signal by the factor $e^{-\lambda\Delta t}$ while a constant background stays unchanged. Fitting the time dependence helps distinguish <radioactive decay> from that background; it does not recover individual capture times.
For an original worked example choose $Rh=20$, $T=W=h$, $d=0$, $\eta=0.8$, $\epsilon=0.5$ and $bh=0.2$. These are illustrative assumptions. Then $N(T)=10/\log 2$, and
$$
\mu_s=\frac2{\log 2}\simeq2.8854,\qquad\mu\simeq3.0854.
$$
Doubling the exposure does not double the surviving inventory: $N(2h)/N(h)=3/2$. A delay of one <half-life> halves the signal. At $\lambda T\ll1$, $N(T)\simeq RT$; at long exposure it saturates at $R/\lambda$. These limits explain why a large target and repeated extraction can matter more than simply waiting longer. The model neglects time-dependent production, finite extraction duration, residual atoms between runs and time-dependent backgrounds.
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