# If the energy of an X-ray photon is "100 keV", then if its wavelength matches that of the ejected core electron, what should be the kinetic energy of that electron?

Aug 14, 2017

About $\text{9.78 keV}$, assuming 100% transfer of energy.

The de Broglie wavelength $\lambda$ is given by

$\lambda = \frac{h}{p} = \frac{h}{m v}$,

where:

• $h = 6.626 \times {10}^{- 34} \text{J"cdot"s}$ is Planck's constant.
• $p = m v$ is the linear momentum in $\text{kg"cdot"m/s}$, for mass $m$ and velocity $v$.

Given the energy of an X-ray photon as $\text{100 keV}$, we have:

${E}_{\text{photon" = "100 keV}} = h \nu = \frac{h c}{\lambda}$

where $\nu$ is the frequency of the photon in ${\text{s}}^{- 1}$ and $c = 2.998 \times {10}^{8} \text{m/s}$ is the speed of light.

And so, we suppose that the photon wavelength is numerically the same as the electron wavelength. I suppose this could occur if one tried to ionize a core electron for X-ray diffraction.

$\lambda = \frac{h c}{E} _ \text{photon} = \frac{h}{p}$

This gives a wavelength of:

$\lambda = \left(6.626 \times {10}^{- 34} \cancel{\text{J"cdotcancel"s" cdot 2.998 xx 10^(8) "m/"cancel"s")/("100 "cancel"k"cancel"eV" xx (1.602 xx 10^(-19) cancel"J")/(cancel"1 eV") xx (1000 cancel"J")/cancel"1 kJ}}\right)$

$= 1.240 \times {10}^{- 11} \text{m}$

Knowing the wavelength, we consider the kinetic energy expressed as a function of linear momentum:

$K = \frac{1}{2} m {v}^{2} = {p}^{2} / \left(2 m\right)$

Thus, with $p = \frac{h}{\lambda}$ from the de Broglie relation, and an electron rest mass of $9.109 \times {10}^{- 31} \text{kg}$, we have that:

$\textcolor{b l u e}{K} = {\left(h / \lambda\right)}^{2} / \left(2 m\right)$

= ((6.626 xx 10^(-34) "kg"cdot"m"^(cancel(2))"/s")/(1.240 xx 10^(-11) cancel"m"))^2/(2 cdot 9.109 xx 10^(-31) cancel"kg")

$=$ $\textcolor{b l u e}{\underline{1.567 \times {10}^{- 15} \text{J}}}$

Or perhaps in more useful units...

1.567 xx 10^(-15) cancel"J" xx (cancel"1 eV")/(1.602 xx 10^(-19) cancel"J") xx ("1 keV")/(1000 cancel"eV")

$=$ $\textcolor{b l u e}{\underline{\text{9.782 keV}}}$

So the X-ray electron has a kinetic energy less than $1 / 10$th of the energy of an X-ray photon, assuming 100% transfer of energy.