The ionization energies of Cobalt (Co),Copper | vedclass

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Question

(B) For the emission of $K$-series characteristic $X$-rays,the energy of the incident electrons $(E = eV)$ must be greater than the ionization energy

Vedclass · Accepted Answer

$K$-series characteristic $X$-rays will be emitted only by $Cu$ and $Co$.

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(B) For the emission of $K$-series characteristic $X$-rays,the energy of the incident electrons $(E = eV)$ must be greater than the ionization energy $(E_i)$ required to knock out an electron from the $K$-shell.Given the operating voltage $V = 15 \, kV$,the energy of the incident electrons is $E = 15 \, keV$.Comparing this with the ionization energies:For $Co$: $7.8 \, keV For $Cu$: $9.0 \, keV For $Mo$: $20.1 \, keV > 15 \, keV$ (Emission not possible).Thus,only $Co$ and $Cu$ can emit $K$-series characteristic $X$-rays.The minimum wavelength of continuous $X$-rays is given by $\lambda_{min} = \frac{hc}{eV}$. Since $V$ is the same for all,$\lambda_{min}$ is the same for all three metals.

The ionization energies of Cobalt $(Co)$,Copper $(Cu)$,and Molybdenum $(Mo)$ are $7.8 \, keV$,$9.0 \, keV$,and $20.1 \, keV$ respectively. If any of these metals are used as a target in an $X$-ray tube operating at $15 \, kV$,then:

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