# 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?

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The **de Broglie wavelength**

#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

#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

#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