INGENIA

RAD-03

Klein–Nishina total snapshot

σ/σ_T ≈ (3/4)[ (1+k)/k³ ((2+2k)/(1+2k) − ln(1+2k)/k) + ln(1+2k)/(2k) − (1+3k)/(1+2k)² ].

Reading speed
AttenuationKlein–Nishina

Governing equation

σKN=σT fKN(k),k=E/mec2\sigma_{KN}=\sigma_T\,f_{KN}(k),\quad k=E/m_ec^2

E∈[1,10000]E \in \left[1, 10000\right]

where

E
Photon energy (keV)
k
E/mc² (—)
\sigma
KN total (b)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

RAD-03 · spectrum
00:0 / 00:08
①

Governing law — red

σKN=σT fKN(k),k=E/mec2\sigma_{KN}=\sigma_T\,f_{KN}(k),\quad k=E/m_ec^2

E∈[1,10000]E \in \left[1, 10000\right]

E=1E = 1k=338.6k = 338.6

②

Domain of the variable — blue

Emin⁡=1,Emax⁡=10000E_{\min} = 1,\quad E_{\max} = 10000

E∈[1,10000]E \in \left[1, 10000\right]

E=1E = 1E=1E = 1

③

Response curve — green

σKN=σT fKN(k),k=E/mec2\sigma_{KN}=\sigma_T\,f_{KN}(k),\quad k=E/m_ec^2

E∈[1,10000]E \in \left[1, 10000\right]

E=1E = 1k=338.6k = 338.6

④

Rate of change — amber

dkdE\frac{\mathrm{d}k}{\mathrm{d}E}

E∈[1,10000]E \in \left[1, 10000\right]

E=63.89E = 63.89k′=−2.678k^{\prime} = -2.678

Lecture brief

Historical brief

Beer attenuation, Compton (1923), Klein–Nishina, Bragg–Gray cavity and KERMA are the transport of photons and charged particles in matter. The sheets compute fluence, kerma and stopping. This sheet (RAD-03 — Klein–Nishina total snapshot) is the form associated with Klein–Nishina. Working symbols: EE →\rightarrow kk, σ\sigma. σ_T = 0.665 b. The cross section falls as ~ (ln k)/k at high energy.

Purpose

Purpose: compute kk, σ\sigma from EE in Radiation physics via σKN=σT fKN(k),k=E/mec2\sigma_{KN}=\sigma_T\,f_{KN}(k),\quad k=E/m_ec^2 σ/σ_T ≈ (3/4)[ (1+k)/k³ ((2+2k)/(1+2k) − ln(1+2k)/k) + ln(1+2k)/(2k) − (1+3k)/(1+2k)² ]. Use it when a real radiation physics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given E=511.000 keVE = 511.000\,\mathrm{keV}, the governing relation σKN=σT fKN(k),k=E/mec2\sigma_{KN}=\sigma_T\,f_{KN}(k),\quad k=E/m_ec^2 yields k=1.000 —k = 1.000\,\mathrm{—}, σ=0.2865 b\sigma = 0.2865\,\mathrm{b}. A σ(E) that falls from Thomson. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Outputs

  • E/mc² k1.000 —
  • KN total \sigma0.2865 b
Reading speed

Watch on YouTube

Free library

Full library

Free PDF / open book

YouTube channels

RAD-03 · spectrum
00:0 / 00:08

Narration of this film

A σ(E) that falls from Thomson.

σ_T = 0.665 b. The cross section falls as ~ (ln k)/k at high energy.

Reading speed

Watch on YouTube