Question

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?

Answer 1

About `"9.78 keV"`, assuming `100%` transfer of energy.

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The de Broglie wavelength `lambda` is given by

`lambda = h/p = h/(mv)`,

where:

• `h = 6.626 xx 10^(-34) "J"cdot"s"` is Planck's constant.
• `p = mv` is the linear momentum in `"kg"cdot"m/s"`, for mass `m` and velocity `v`.

Given the energy of an X-ray photon as `"100 keV"`, we have:

`E_"photon" = "100 keV" = hnu = (hc)/lambda`

where `nu` is the frequency of the photon in `"s"^(-1)` and `c = 2.998 xx 10^(8) "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 = (hc)/E_"photon" = h/(p)`

This gives a wavelength of:

`lambda = (6.626 xx 10^(-34) cancel"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")`

`= 1.240 xx 10^(-11) "m"`

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

`K = 1/2 mv^2 = p^2/(2m)`

Thus, with `p = h/lambda` from the de Broglie relation, and an electron rest mass of `9.109 xx 10^(-31) "kg"`, we have that:

`color(blue)(K) = (h//lambda)^2/(2m)`

`= ((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")`

`=` `color(blue)ul(1.567 xx 10^(-15) "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")`

`=` `color(blue)ul("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.