RPR-28

Relaxation length

λ = 1/μ, I = I0 e^{−x/λ}. The 1/e attenuation length.

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ShieldingRelaxation length

Governing equation

\lambda=1/\mu,\quad I=I_0 e^{-x/\lambda}

where

\mu: Attenuation μ (1/cm)
I_0: Entrance rate (µSv/h)
x: Thickness (cm)
\lambda: Relaxation length (cm)
I: Transmitted rate (µSv/h)

Lecture brief

Historical brief

Inverse-square, half-value layer, ICRP weighting and ALARA are the protection craft since the 1920s commissions. The lab computes transmission, equivalent dose and a shielding snapshot. This sheet (RPR-28 — Relaxation length) is the form associated with Relaxation length. Working symbols: $\mu$ , $I_0$ , $x$ $\rightarrow$ $\lambda$ , $I$ . HVL = λ ln 2, TVL = λ ln 10. Concrete λ is a few cm at diagnostic kV and tens of cm at MV.

Purpose

Purpose: compute

\lambda

I

from

\mu

I_0

x

in Radiation protection via

\lambda=1/\mu,\quad I=I_0 e^{-x/\lambda}

λ = 1/μ, I = I0 e^{−x/λ}. The 1/e attenuation length. Use it when a real radiation protection question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

\mu = 0.150\,\mathrm{1/cm}

I_0 = 50.000\,\mathrm{\mu Sv/h}

x = 30.000\,\mathrm{cm}

, the governing relation

\lambda=1/\mu,\quad I=I_0 e^{-x/\lambda}

yields

\lambda = 6.667\,\mathrm{cm}

I = 0.555\,\mathrm{\mu Sv/h}

. An exponential fading through a thick wall. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Attenuation μ \mu0.15 1/cmEntrance rate I_050 µSv/hThickness x30 cm

Outputs

Relaxation length \lambda6.667 cm
Transmitted rate I0.555 µSv/h

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Narration of this film

An exponential fading through a thick wall.

HVL = λ ln 2, TVL = λ ln 10. Concrete λ is a few cm at diagnostic kV and tens of cm at MV.

Reading speed

Watch on YouTube