Kramers' law

Kramers' law is a formula for the spectral distribution of X-rays produced by an electron hitting a solid target. The formula concerns only bremsstrahlung radiation, not the element specific characteristic radiation. It is named after its discoverer, the Dutch physicist Hendrik Anthony Kramers.[1]

The formula for Kramers' law is usually given as the distribution of intensity (photon count) ${\displaystyle I}$ against the wavelength ${\displaystyle \lambda }$ of the emitted radiation:[2]

${\displaystyle dI(\lambda )=K\left({\frac {\lambda }{\lambda _{min}}}-1\right){\frac {1}{\lambda ^{2}}}d\lambda }$

The constant K is proportional to the atomic number of the target element, and ${\displaystyle \lambda _{min}}$ is the minimum wavelength given by the Duane–Hunt law. The maximum intensity is ${\displaystyle {\frac {K}{4\lambda _{min}^{2}}}d\lambda }$ at ${\displaystyle 2\lambda _{min}}$.

The intensity described above is a particle flux and not an energy flux as can be seen by the fact that the integral over values from ${\displaystyle \lambda _{min}}$ to ${\displaystyle \infty }$ is infinite. However, the integral of the energy flux is finite.

To obtain a simple expression for the energy flux, first change variables from ${\displaystyle \lambda }$ (the wavelength) to ${\displaystyle \omega }$ (the angular frequency) using ${\displaystyle \lambda =2\pi c/\omega }$ and also using ${\displaystyle {\tilde {I}}(\omega )=I(\lambda ){\frac {-d\lambda }{d\omega }}}$. Now ${\displaystyle {\tilde {I}}(\omega )}$ is that quantity which is integrated over ${\displaystyle \omega }$ from 0 to ${\displaystyle \omega _{max}}$ to get the total number (still infinite) of photons, where ${\displaystyle \omega _{max}=2\pi c/\lambda _{min}}$:

${\displaystyle {\tilde {I}}(\omega )={\frac {K}{2\pi c}}\left({\frac {\omega _{max}}{\omega }}-1\right)}$

The energy flux, which we will call ${\displaystyle \psi (\omega )}$ (but which may also be referred to as the "intensity" in conflict with the above name of ${\displaystyle I(\lambda )}$) is obtained by multiplying the above ${\displaystyle {\tilde {I}}}$ by the energy ${\displaystyle \hbar \omega }$:

${\displaystyle \psi (\omega )={\frac {K}{2\pi c}}(\hbar \omega _{max}-\hbar \omega )}$

for ${\displaystyle \omega \leq \omega _{max}}$

${\displaystyle \psi (\omega )=0}$

for ${\displaystyle \omega \geq \omega _{max}}$.

It is a linear function that is zero at the maximum energy ${\displaystyle \hbar \omega _{max}}$.