NUC-09

Bethe–Bloch stopping

−dE/dx = K z² (Z/A) (1/β²) [ln(2 m_e c² β² γ² / I) − β²]. Heavy charged particle.

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NuclearBethe–Bloch

Governing equation

-\dfrac{dE}{dx}=K z^2\dfrac{Z}{A}\dfrac1{\beta^2}\left[\ln\dfrac{2m_ec^2\beta^2\gamma^2}{I}-\beta^2\right]

$\beta \in \left[0.02, 0.95\right]$

where

z: Projectile charge (—)
Z: Target Z (—)
A: Target A (—)
\beta: v/c (—)
-dE/dx: Stopping (MeV cm²/g)

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.

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①

Governing law — red

-\dfrac{dE}{dx}=K z^2\dfrac{Z}{A}\dfrac1{\beta^2}\left[\ln\dfrac{2m_ec^2\beta^2\gamma^2}{I}-\beta^2\right]

$\beta \in \left[0.02, 0.95\right]$

$\beta = 0.02$ $-dE/dx = 682.7$

Along the curve

②

Domain of the variable — blue

\beta_{\min} = 0.02,\quad \beta_{\max} = 0.95

$\beta \in \left[0.02, 0.95\right]$

$\beta = 0.02$ $\beta = 0.02$

Along the curve

③

Response curve — green

-\dfrac{dE}{dx}=K z^2\dfrac{Z}{A}\dfrac1{\beta^2}\left[\ln\dfrac{2m_ec^2\beta^2\gamma^2}{I}-\beta^2\right]

$\beta \in \left[0.02, 0.95\right]$

$\beta = 0.02$ $-dE/dx = 682.7$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}-dE/dx}{\mathrm{d}\beta}

$\beta \in \left[0.02, 0.95\right]$

$\beta = 0.026$ $-dE/dx^{\prime} = -23100.2$

Along the curve

Lecture brief

Historical brief

Rutherford, Chadwick, the semi-empirical mass formula, fission (Hahn–Strassmann 1938) and fusion Q-values, then Compton and Bethe–Bloch stopping, are the nuclear toolkit. Decay, binding and dose start here. This sheet (NUC-09 — Bethe–Bloch stopping) is the form associated with Bethe–Bloch. Working symbols: $z$ , $Z$ , $A$ , $\beta$ $\rightarrow$ $-dE/dx$ . K = 0.307 MeV cm²/g. I ≈ 11.5 Z eV as a pedagogical mean ionisation.

Purpose

Purpose: compute

-dE/dx

from

z

Z

A

\beta

in Nuclear physics via

-\dfrac{dE}{dx}=K z^2\dfrac{Z}{A}\dfrac1{\beta^2}\left[\ln\dfrac{2m_ec^2\beta^2\gamma^2}{I}-\beta^2\right]

−dE/dx = K z² (Z/A) (1/β²) [ln(2 m_e c² β² γ² / I) − β²]. Heavy charged particle. Use it when a real nuclear physics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

z = 1.000\,\mathrm{—}

Z = 6.000\,\mathrm{—}

A = 12.000\,\mathrm{—}

\beta = 0.150\,\mathrm{—}

, the governing relation

-\dfrac{dE}{dx}=K z^2\dfrac{Z}{A}\dfrac1{\beta^2}\left[\ln\dfrac{2m_ec^2\beta^2\gamma^2}{I}-\beta^2\right]

yields

-dE/dx = 39.63\,\mathrm{MeV cm^{2}/g}

. A track, a dE/dx that rises toward a Bragg peak. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Projectile charge z1 —Target Z Z6 —Target A A12 —v/c \beta0.15 —

Outputs

Stopping -dE/dx39.63 MeV cm²/g

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Narration of this film

A track, a dE/dx that rises toward a Bragg peak.

K = 0.307 MeV cm²/g. I ≈ 11.5 Z eV as a pedagogical mean ionisation.

Reading speed

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