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$I(x) = I_0 \cdot e^{~-u \cdot x}$

Where u is the linear attenuation coefficient

And how does this relate to the following

$N(x) = N_0 \cdot e^{~-u \cdot x}$

Where N is the count rate of the beam typically measured by a GM counter.

Andreas
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    I suggest that you provide more background about why you think that the formula applies to the gamma rays, what specific situation you are talking about, and what $N(x)$ is in your case. – Roger V. Dec 03 '20 at 14:20

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A gamma photon is either absorbed or carries on through matter without being changed. This implies that the probability, $K$, per unit distance travelled of a gamma being absorbed is independent of how far the gamma has travelled. So we have $$\frac{(–dN)/N}{dx}=K \ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \frac{dN}{dx}=\ –KN$$ Integration leads to the equation you have quoted.

Alphas and betas lose their kinetic energy bit by bit is a series of ionisation events, so the further they have travelled, the slower they get. Eventually they stop. This finite range is quite unlike the exponential fall-off in gamma intensity with distance.

Philip Wood
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  • Ah, thanks for clearing that up. By typing the bottom equation with N instead of I, I was also hoping that someone could explain to me what the reasoning is for the Intensity being directly proportional to the count rate. Or, in other words. I = k*N – Andreas Dec 03 '20 at 14:47
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    You might define intensity as number of particles crossing unit normal area per second. So if your Geiger-Müller detects a fixed fraction of the gammas that enter it, the count rate will be proportional to the beam intensity. The proportionality will still hold (for a given energy of gamma) if you define beam intensity as energy per second per unit normal area. – Philip Wood Dec 03 '20 at 14:53
  • In addition, charged particles experience energy loss from both electronic and nuclear stopping, and the relative amounts change with the energy of the particle. – Jon Custer Dec 03 '20 at 15:32
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Why does the formula for decreasing intensity of radiation only work for gamma rays?

Incorrect. Beer–Lambert law works for many radiation types, including photons (gamma rays, visible light, ...), neutrons, etc. General form of Lambert law sounds : $$ A = \varepsilon \ell c $$ Where $A$ is absorbance, $\varepsilon$ is the molar attenuation coefficient, $\ell$ is the radiation path length, and $c$ is is the concentration of the attenuating species.

Absorbance can be defined as : $$ A=\ln\left(\frac {I_0}{I}\right) $$ Where $I_0$ is incident ray intensity, $I$ - transmitted ray intensity. Substituting this into equation above and using logarithm rules for expressing logarithm argument and making a couple adjustments, gives : $$ I=I_0 e^{-\varepsilon~ \ell ~c} $$

This is in case attenuation coefficient is uniform in all media, otherwise one needs to integrate with respect to $\varepsilon(\ell)$.

Agnius Vasiliauskas
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