The K_{alpha} X-ray emission from tungsten oc | vedclass

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(C) The energy difference between the $K$ and $L$ levels is given by the energy of the emitted $K_{\alpha}$ photon: $E = \frac{hc}{\lambda}$ Substitut

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$59 \ keV$

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(C) The energy difference between the $K$ and $L$ levels is given by the energy of the emitted $K_{\alpha}$ photon: $E = \frac{hc}{\lambda}$ Substituting the values: $E = \frac{(6.63 	imes 10^{-34} \ J \cdot s) 	imes (3 	imes 10^8 \ m/s)}{0.021 	imes 10^{-9} \ m}$ $E \approx 9.47 	imes 10^{-15} \ J$ To convert this energy into electron-volts $(eV)$,divide by $1.6 	imes 10^{-19} \ J/eV$: $E = \frac{9.47 	imes 10^{-15}}{1.6 	imes 10^{-19}} \ eV \approx 59,187 \ eV$ $E \approx 59 \ keV$.

The $K_{\alpha}$ $X$-ray emission from tungsten occurs at $\lambda = 0.021 \ nm$. The energy difference between the $K$ and $L$ levels in the atom is:

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