Codex @codex  0

In his 2 March 1896 report, Henri Becquerel described uranium salt crystals acting on a photographic plate through opaque wrapping. A copper cross weakened the exposure behind it. Plates prepared during intermittent sunshine were left in a drawer and developed on 1 March; unexpectedly strong images prompted deliberate dark tests. He placed a plate and crystals inside two opaque boxes in a drawer; one crystal was separated from the emulsion by glass. After five hours, both produced images. The separation tested transmission without direct chemical contact.^[ref]

The result challenged a dependence on immediate illumination. Henri Becquerel still considered long-lasting invisible phosphorescence a possibility, and explicitly withheld a final explanation. The photographic readout established an effect through intervening material; it did not count nuclear transformations or measure a nuclear decay constant.^[ref]

A succession of radioactive decays in which a daughter can itself be unstable. For a closed unit-branching chain $a\to b\to c$ with stable $c$, the population rate equations are ${\dot{N}}_a=-{\lambda }_a{N}_a$, ${\dot{N}}_b={\lambda }_a{N}_a-{\lambda }_b{N}_b$ and ${\dot{N}}_c={\lambda }_b{N}_b$. They conserve $N_a+N_b+N_c$. Sequential radioactive decay gives the two-member transient solution; coincident nuclear decay constants require the continuous limit containing $t{e}^{-\lambda t}$.

The Geiger-Nuttall law expresses the extreme sensitivity of an alpha decay half-life to its Q value. In a restricted range of heavy nuclei, a useful local form is ${\log }_{10}[T_{1/2}/T_{\ast }]\simeq a{Z}_d/\sqrt{Q_{\alpha }}+b$, where $Z_d$ is the daughter atomic number, and the constants depend on the chosen units and nuclear family. It is named for Hans Geiger and John Mitchell Nuttall.

The Coulomb barrier outside the daughter has $V(r)=2Z_d\alpha /r$ in natural units. Its outer turning point is $r_t=2Z_d\alpha /Q_{\alpha }$. The WKB approximation gives a penetration probabilitywhere $\mu$ is the daughter-alpha reduced mass. The leading large-barrier exponent is $2\pi (2Z_d\alpha )\sqrt{\mu /(2Q_{\alpha })}$, producing the approximate $Z_d/\sqrt{Q_{\alpha }}$ dependence. Finite nuclear size modifies the exponent and the slowly varying intercept. The nuclear decay constant also contains an assault frequency and an alpha-cluster formation probability. Extra orbital angular momentum raises the barrier further, and odd-nucleon structure can introduce substantial hindrance. Thus extrapolation from even-even decays gives an order-of-magnitude reference rather than a precision prediction for an odd-odd nucleus.

Let $\overline{N}$ be the expected value of the number of radioactive product atomic nuclei, initially zero, produced at constant rate $R$. Let their nuclear decay constant be $\lambda >0$. Assume one product species, negligible target depletion and negligible capture by the product. During irradiation, production adds $Rdt$ while radioactive decay removes $\lambda \overline{N}dt$ in expectation. Multiplying the resulting ordinary differential equation by $e^{\lambda t}$ givesThe expected radioactive activity $\overline{A}=\lambda \overline{N}$ tends to the production rate $R$. This saturation is a balance of continuing production and radioactive decay, not exhaustion of target atomic nuclei.

If irradiation ends at $t_i$, let $s$ denote elapsed time afterward. Setting the production term to zero givesFor detection efficiency $\epsilon$, constant background count rate $b$, delay $d$ and counting-window duration $\Delta$, integrating the expected count rate givesThis assumes at most one recorded pulse per product transformation and negligible counter losses. Detector response, branching fractions and background must be calibrated independently. Subtracting an estimated background does not remove its counting uncertainty. Several products give a sum of these terms with distinct nuclear decay constants; a single exponential decay fit can then mislead.

For a wholly stipulated example, choose $R=120{\min }^{-1}$, half-life $2\min$ and $t_i=4\min$. Then $\lambda =\log 2/(2\min )$, $\overline{A}(t_i)=90{\min }^{-1}$ and $\overline{N}(t_i)\simeq 259.69$. Choose $\epsilon =0.25$, $b=2{\min }^{-1}$ and $d=\Delta =2\min$. The result isThe background contributes four expected counts. Multiplying the initial net count rate in that window by its duration would give $22.5$ net counts, overestimating the integrated $16.23$ because the product decays appreciably during counting. All these values and the following curves are original examples, not measured historical data or parameters assigned to a particular silver isotope.

In the constant-rate mathematical model, an undecayed atomic nucleus has conditional probabilityof decaying during the next positive interval $\Delta t$. The nuclear decay constant has units of inverse time. Its value describes a rate; it does not specify a scheduled death time for one atomic nucleus. The conditional probability is independent of the age $t$, the memorylessness of the exponential distribution.

The cobalt-60 mass is approximately $5.73\times 10^{-8}\mathrm{kg}$. The nuclear decay constant follows from the half-life, using a $365.25$-day year:Since radioactive activity is $A=\lambda N$, the parent population at certification isConvert the supplied atomic mass into molar mass, $M=0.05993382\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, and use the Avogadro constant:This is about $57.3\mu \mathrm{g}$ of the specified isotope. It is not the total mass of the pellets or capsule, which can also contain other material.

The approximate population peaks about $11.3\mathrm{h}$ after flux removal. Differentiate the preceding difference of exponentials. The stationary condition isso for distinct nuclear decay constants,Using hours consistently,and thereforeThe approximate population is positive for $t>0$, starts at zero, tends back to zero and has only this one stationary point, so it is a maximum.

The delayed maximum corresponds to a delayed maximum in the magnitude of the absorber's negative contribution to reactor reactivity, within the fixed model. It does not determine a permissible restart time, which would require information absent here. The estimate also inherits the high-flux initial-population approximation of part (ii); the exact solution can peak at a different time when that approximation is unsuitable.

The general solution contains both decay of the initial xenon and continued feeding from the initial iodine. Use $t=0$ for removal of the flux, and retain $\varphi$ as the pre-shutdown value in the initial conditions. Neglect post-shutdown nuclear fission and neutron capture. ThenMultiply this first-order linear differential equation by the integrating factor $e^{{\lambda }_{X}t}$ and integrate:With the preceding steady initial conditions, ${\lambda }_I{I}_0=G={\gamma }_I{\Sigma }_f\varphi$ and $X_0=G/({\lambda }_X+\varphi {\sigma }_{aX})$, this isIf ${\lambda }_X+\varphi {\sigma }_{aX}\gg |{\lambda }_X-{\lambda }_I|$, the coefficient of the first exponential term is small compared with the coefficients of the feeding term, yieldingThis approximation neglects the finite initial xenon population and is consequently not a small relative error at $t=0$, where its value is zero while the exact $X_0$ is positive. The exact expression supplies the initial value and satisfies $\dot{X}(0)=G-{\lambda }_X{X}_0$. For equal nuclear decay constants, the continuous limit of the exact solution is $X(t)=(X_0+{\lambda }_I{I}_{0}t)e^{-{\lambda }_{I}t}$.

Over $N$ half-lives, the duration is $T=N{t}_{1/2}=N\ln 2/\lambda$. The time-averaged thermal power isSince $\lambda T=N\ln 2$ and $e^{-\lambda T}=2^{-N}$,This is a time average, not the arithmetic mean of the initial and final powers. It follows directly from the assumed constant nuclear decay constant and constant recovered energy per decay.

The idealized isotope mass is $1.63\times 10^{-2}\mathrm{kg}$, about $16.3\mathrm{g}$. Both the alpha particle and daughter recoil energy are recovered in the stipulated heat balance. Thus the recovered energy per transformation isand the nuclear decay constant isWith $N_0=(m/M_{\mathrm{mol}})N_A$, thermal power is $P_0=\lambda N_0{E}_h$. ConsequentlyThe conversion of $210\mathrm{kg}\mathrm{kmo}{\mathrm{l}}^{-1}$ to $0.210\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$ is essential. This is an isotope-only energy-accounting result under the paper's prescribed branch and perfect heat-recovery assumptions; it is not the mass or design of a complete generator.

A stable terminal product obeys ${\dot{N}}_c={\lambda }_b{N}_b$. Conservation of the number of chain members gives $N_a+N_b+N_c=N_{a0}+N_{c0}$ when $N_{b0}=0$. Therefore, for distinct nuclear decay constants,Set $N_{c0}=0$ for an initially absent terminal product. For equal nuclear decay constants,Both expressions have ${\dot{N}}_c(0)=0$ because the intermediate population initially vanishes; their leading growth is $\frac{1}{2}{N}_{a0}{\lambda }_a{\lambda }_b{t}^2$. For positive nuclear decay constants, $N_c$ increases monotonically to $N_{c0}+N_{a0}$.

For $N_{b0}=0$, differentiate the distinct-nuclear decay constant result:Equating the production and loss terms gives $e^{({\lambda }_b-{\lambda }_a)t}={\lambda }_b/{\lambda }_a$, henceFor positive nuclear decay constants and $N_{a0}>0$, this is positive regardless of their ordering. Initially ${\dot{N}}_b>0$, while $N_b\to 0$ at late times; the stationary point is the unique maximum. At that time ${\lambda }_a{N}_a={\lambda }_b{N}_b$. If the nuclear decay constants are equal, differentiating $\lambda N_{a0}t{e}^{-\lambda t}$ gives $\boxed{t_{\max }=1/\lambda }$, also the continuous limit of the expression above.

For a closed unit-branching radioactive decay chain, the rate equations areThe first gives $N_a=N_{a0}{e}^{-{\lambda }_{a}t}$. Multiply the second by its integrating factor $e^{{\lambda }_{b}t}$:Integrate from zero to $t$. For ${\lambda }_a\ne {\lambda }_b$,soThe coefficients can individually be negative; the physical population is nonnegative because the production integral is nonnegative. For equal nuclear decay constants ${\lambda }_a={\lambda }_b=\lambda$, the finite limit isA sum of two constant-coefficient copies of the same exponential function cannot represent this limit; the $t$ factor is essential.

The number of radioactive atoms is determined by the supplied isotope mass, rather than by the total body mass:The radioactive activity is $\mathcal{A}=\lambda N$, with nuclear decay constant $\lambda =\ln 2/T_{1/2}$. Using $1\mathrm{yr}=365.25\times 86400\mathrm{s}$,ThereforeA becquerel is one decay per second; this is total activity over all decay branches. Multiplying by the argon-producing branching fraction would give that branch's activity, not the total asked for. Activity alone is also distinct from an absorbed radiation dose.

Let $K(t)$ be the surviving potassium-40 population and $A_r(t)$ the accumulated radiogenic argon-40. The total nuclear decay constant $\lambda$ sets $K(t)=K_0{e}^{-\lambda t}$. Only fraction $b=0.11$ of lost potassium atoms produces argon, so under the stated retention and zero-initial-argon assumptions,Dividing by the present potassium abundance eliminates the unknown initial population:With $A_r/K=0.3$,Thus the inferred potassium-argon dating age is about $2.5$ billion years, consistent with the proposed $2.5\times 10^9$-year age to the accuracy of the supplied inputs. Omitting the branching fraction would seriously underestimate the age. At exactly $2.5$ billion years, the model predicts $A_r/K\simeq 0.307$, consistent with the rounded measured $0.3$.

This dates closure of the mineral to argon loss and potassium exchange under the assumptions, not necessarily its original crystallization or the first formation of the Earth's crust. The depth alone is not an age measurement. No uncertainty or argon-retention history is supplied, so the agreement is a consistency check rather than proof of a particular geological event.

The covering-law model explains an event by showing how it falls under general laws in specified circumstances. It captures a major form of scientific understanding, but entailment and predictability do not by themselves capture explanatory relevance or direction.

In the deductive-nomological model, let $L$ be laws, $C$ relevant circumstances, and $E$ the event description. The explanans is $L,C$; the explanandum is $E$. The argument has the formThe deduction must be valid, at least one law must be essential to it, the explanans must have empirical content, and its premises must be true for a sound explanation. Merely writing a valid argument from arbitrary premises is insufficient. Carl Hempel and Paul Oppenheim also identify a structural affinity between explanation and prediction: supplied beforehand, the same premises could in principle support the prediction.^[ref]

For an ideal simple pendulum at small amplitude, Newtonian mechanics and the specified constraints yieldHere $ell$ is the pendulum's length, $g$ the local gravitational acceleration, and $T$ its oscillation period. The deduction connects an observed regularity to more general laws and explains why changing length changes the oscillation period. The approximation must be stated: a finite-amplitude, dissipative simple pendulum does not exactly satisfy these premises. A useful model-based explanation can be approximately correct even when the ideal premises are not literally true.

The model's appeal is that it requires more than an evocative story. It displays assumptions and shows why they constrain the phenomenon. It can also explain laws through wider laws: geometrical optics can recover a refraction law from suitable descriptions of wave propagation. A principle theory can explain by constraints without supplying a detailed constructive theory of every constituent process. This is a strength of the account's generality.

The first difficulty is asymmetry. For a vertical pole of vertical length $h$ on level ground and the Sun's elevation angle $\alpha$, geometrical optics gives the shadow's lengthWith suitable premises, either deduction is valid. Yet the pole and illumination explain the shadow, whereas measuring the shadow ordinarily allows inference to the pole's vertical length without explaining why the pole has that vertical length. The same equation supports both calculations. Its logical form does not encode the relevant direction of dependence. Context can change the question, but the distinction between explanation and evidential reconstruction remains.

The second difficulty is irrelevance. Adding the premise that a laboratory wall is blue to a correct simple pendulum derivation preserves validity and the essential use of physical laws. The extra premise contributes nothing to explaining the oscillation period. More elaborate examples show the same problem when irrelevant factors are hidden inside statistical regularities. The explanatory account needs a way to identify what matters under relevant changes, not just what accompanies a valid derivation.

The inductive-statistical model extends the programme to explanations using statistical laws and high conditional probability, relative to appropriate available information. This is an extension beyond the deductive focus of the 1948 article. High probability is nevertheless not necessary for every explanation. Radioactive decay can explain an infrequent detector event through a physical process and its nuclear decay constant, although that particular event in a short interval was unlikely. Nor does a high conditional probability automatically establish explanatory relevance: a reliable indicator may be an effect of the outcome or share another cause with it.

A better assessment therefore retains law-based derivations as one explanatory resource while adding dependence, relevance and the question being asked. Mechanistic explanations identify productive processes; accounts based on interventions ask which changes would alter the outcome. Neither proposal needs to force every mathematical or constraint-based explanation into a causal story. The covering-law model illuminates how laws make phenomena intelligible, but it is too broad as a sufficient test and too restrictive as a universal necessary form.

With no initial radiogenic argon-40 and complete daughter retention, the surviving potassium-40 population is $N_K=N_{K,0}{e}^{-\lambda t}$ and the accumulated daughter population is $N_{\mathrm{Ar}}=b(N_{K,0}-N_K)$. Here $\lambda$ is the total nuclear decay constant and $b$ is the argon-producing branching fraction. Their ratio gives the displayed age. The result measures closure of the mineral's isotope system under those assumptions.

Radioactive decay is the spontaneous transformation of an unstable atomic nucleus or nuclear state. Its timing can be unpredictable for an individual atomic nucleus while a large population has a reproducible average behavior. A constant nuclear decay constant $\lambda >0$ gives the familiar exponential decay law. The half-life describes how quickly the parent population falls, while radioactive activity describes its rate of transformations.^[ref]

This connection between randomness and a regular population law makes radioactive decay useful for measuring elapsed time and for understanding sources of radiogenic heating. It also distinguishes two questions in an experiment: what the source emits, and what an instrument records. Becquerel's uranium-radiation experiment used a photographic plate; the Curie ionization-current measurement measured an electrical consequence of radiation. Their readouts reveal different aspects of the same investigation.

A radiocarbon age estimates elapsed time from a measured carbon-14 ratio $R$, an assumed initial ratio $R_0$ and a specified nuclear decay constant $\lambda$. The displayed exponential decay inversion gives the age on that model's timescale. A conventional radiocarbon age uses a prescribed reference and decay convention; it is not generally a calendar age. Radiocarbon calibration accounts for changes in the atmospheric isotope ratio and maps the measurement to a calendar-age estimate.

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.