A gas-filled particle detector using electron avalanche multiplication. Within its proportional operating range, the collected electric charge tracks the initial ionization, allowing more than a yes-or-no count.

An instrument that records a time-varying electrical signal.

A collector that measures incident charged-particle electric charge.

A load cell is a calibrated device for measuring force. In a tensile modulus experiment its output supplies the axial load used to calculate engineering stress.

Let $N_{\mathrm{inc}}$ be the number of projectiles actually incident on the foil during an exposure, $n_s$ its foil areal number density, and $\eta$ the probability of detecting an accepted particle. For a sufficiently small detector patch of area $A$ at distance $R$ from the scattering point, with its normal at angle $\psi$ to the line of sight,The product $n_s(d\sigma /d\Omega )\Delta \Omega$ is a dimensionless scattering probability in this approximation; $\mu$ is an expected value of counts, so it need not be an integer. The approximation requires a nearly constant differential scattering cross-section and efficiency across the patch and incident beam. For finite acceptance use $\mu =N_{\mathrm{inc}}{n}_s{\int }_{\mathrm{detector}}\eta (\Omega )(d\sigma /d\Omega )d\Omega$, including the incident beam's spread if appreciable. A screen facing the foil has $\psi =0$.

For an original numerical model, stipulate monoenergetic alpha particles with $z=2$, $E=5.00\mathrm{MeV}$, and a gold target with $Z=79$. Use the rounded model constant $e^2/(4\pi {\epsilon }_0)=1.440\mathrm{MeV}\mathrm{fm}$, where $1\mathrm{MeV}=10^6$ electronvolts and $1\mathrm{fm}=10^{-15}\mathrm{m}$. Thus $\kappa =227.52\mathrm{MeV}\mathrm{fm}$ and $a=22.752\mathrm{fm}$. Choose $N_{\mathrm{inc}}=10^{11}$, $n_s=10^{21}{\mathrm{m}}^{-2}$, $\eta =1$, and $\Delta \Omega =10^{-3}\mathrm{sr}$. For example, $A=1\mathrm{m}{\mathrm{m}}^2$ facing the foil at $R\simeq 31.6\mathrm{mm}$ gives this small-patch solid angle.

At $\theta =90^{\circ }$, $b=a$ and $d\sigma /d\Omega =a^2=517.654\mathrm{f}{\mathrm{m}}^2\mathrm{s}{\mathrm{r}}^{-1}=5.17654\mathrm{b}\mathrm{s}{\mathrm{r}}^{-1}$, using $1$ barn $=100\mathrm{f}{\mathrm{m}}^2=10^{-28}{\mathrm{m}}^2$. The expected signal is $\mu \simeq 51.8$ counts. At $60^{\circ }$ it is four times larger, about $207$ counts. Doubling foil areal number density or detector solid angle doubles these expected signals as long as the approximations continue to hold.

An observation also contains backgrounds. For a stable source and matched geometry, an estimator is $C_{\mathrm{net}}=C_{\mathrm{foil}}-(T_{\mathrm{foil}}/T_{\mathrm{blank}})C_{\mathrm{blank}}$, where $T$ denotes exposure duration. If source activity changes, normalize by incident exposure as well. Background subtraction removes an estimated background; it does not make finite counting fluctuations disappear.

For an incident beam uniform across its relevant impact parameters, an incident annulus has area $d\sigma =2\pi b|db|$. Rotational symmetry maps it to outgoing solid angle $d\Omega =2\pi \sin \theta d\theta$. The central-potential classical differential cross-section and the Rutherford impact-parameter relation therefore giveThis is a differential scattering cross-section, with units of area per steradian, conventionally ${\mathrm{m}}^2\mathrm{s}{\mathrm{r}}^{-1}$, not a detector count and not an area per degree. It applies for $0<\theta <\pi$, with a finite limiting value at $\pi$. Doubling incident kinetic energy divides it by four at fixed angle; doubling either electric charge number multiplies it by four.

A useful independent check integrates over all azimuths between two positive angles. For $0<{\theta }_1<{\theta }_2\le \pi$,This equals the incident annulus area $\pi (b_1^2-b_2^2)$ directly. A detector covering only a small patch at one angle does not collect this full azimuthal band.

Let $\theta$ be the scattering angle between incoming and outgoing velocities, so forward scattering is $\theta =0$ and exact reversal is $\theta =\pi$. Define the length $a=\kappa /(2E)$. ThenFor $b>0$, this gives $0<\theta <\pi$; $b=0$ is the head-on limiting case $\theta =\pi$. Large deflections select small impact parameters, rather than requiring every incident alpha particle to reverse.

A short derivation uses conservation of angular momentum and conservation of energy. Choose the polar angle $\varphi =0$ at closest approach, and write $u=1/r$. The Binet equation for the repulsive Coulomb potential givesThe last equality follows by substituting this orbit into $E=m{\dot{r}}^2/2+L^2/(2m{r}^2)+\kappa /r$. The hyperbola reaches infinity at $\varphi =\pm {\varphi }_{\infty }$, with $\cos {\varphi }_{\infty }=1/ϵ$. Its scattering angle is $\theta =\pi -2{\varphi }_{\infty }$, henceThis proves the relation without assuming that the deflection is small. The symbol $ϵ$ here is a dimensionless orbital parameter, distinct from ${\epsilon }_0$.

For a head-on repulsive Rutherford scattering encounter, relative kinetic energy becomes Coulomb potential energy at closest approach. With projectile charge $ze$ and target charge $Ze$, conservation of energy gives the displayed separation. If the Coulomb description remains valid down to that separation, it bounds the target's interaction radius from above, treating the projectile as pointlike. For a stationary target of mass $M$ and incident projectile mass $m$, use centre-of-mass energy $E_{\mathrm{cm}}=M{E}_{\mathrm{lab}}/(M+m)$; neglecting nuclear recoil amounts to $M\gg m$.

By Rüdiger Paschotta. No article date is established.

Online text: www.rp-photonics.com/faraday_effect.html

Article by Rüdiger Paschotta in RP Photonics Encyclopedia by Rüdiger Paschotta, hosted by RP Photonics. No article publication or revision date is established.

Online text: www.rp-photonics.com/refractive_index.html

Cited passages: What is a Refractive Index?; Wavelength Dependence and Other Dependencies; Anisotropic Media; Complex Refractive Index; Group Index and Effective Index; Measuring Refractive Indices.