The K_{alpha} X-ray of molybdenum has a wavel | vedclass

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(D) The energy of the $K_{\alpha}$ $X$-ray photon is given by the difference in energy between the $K$ shell and the $L$ shell: $E_{K_{\alpha}} = E_{K

Vedclass · Accepted Answer

$10$

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(D) The energy of the $K_{\alpha}$ $X$-ray photon is given by the difference in energy between the $K$ shell and the $L$ shell: $E_{K_{\alpha}} = E_{K} - E_{L}$.The energy of the photon is calculated as $E_{K_{\alpha}} = \frac{hc}{\lambda}$.Given $h = 4.14 	imes 10^{-15} \, eVs$ and $c = 3 	imes 10^{8} \, ms^{-1}$,the product $hc = 12.42 	imes 10^{-7} \, eV \cdot m$.Substituting the wavelength $\lambda = 0.071 	imes 10^{-9} \, m$:$E_{K_{\alpha}} = \frac{12.42 	imes 10^{-7}}{0.071 	imes 10^{-9}} \, eV \approx 17493 \, eV \approx 17.5 \, keV$.We are given the energy of the atom with a $K$ electron removed as $E_{K} = 27.5 \, keV$.Using the relation $E_{L} = E_{K} - E_{K_{\alpha}}$:$E_{L} = 27.5 \, keV - 17.5 \, keV = 10 \, keV$.

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