Radiochemical neutrino counting model (${\mu }_s=ϵ\eta \frac{R}{\lambda }(1-e^{-\lambda T})e^{-\lambda d}(1-e^{-\lambda W})$)

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 obeysThe 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, $ϵ$ the probability of recording a decay, and $W$ the counting duration. The expected signal pulses areWith constant background rate $b$, the expected total is $\mu ={\mu }_s+bW$. During counting, the expected instantaneous pulse rate is $s(t)+b$, with $s(t)=ϵ\eta \lambda N(T)e^{-\lambda (d+t)}$. 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 $Rh=20$, $T=W=h$, $d=0$, $\eta =0.8$, $ϵ=0.5$ and $bh=0.2$. These are illustrative assumptions. Then $N(T)=10/\log 2$, andDoubling the exposure does not double the surviving inventory: $N(2h)/N(h)=3/2$. A delay of one half-life halves the signal. At $\lambda T\ll 1$, $N(T)\simeq RT$; 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.