The K_alpha X-ray emission line for tungsten | vedclass

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(B) The energy of the emitted $K_\alpha$ photon is given by the difference in energy between the $K$ and $L$ shells: $\Delta E = E_K - E_L = \frac{hc}

Vedclass · Accepted Answer

$59$

Vedclass · Answer

(B) The energy of the emitted $K_\alpha$ photon is given by the difference in energy between the $K$ and $L$ shells: $\Delta E = E_K - E_L = \frac{hc}{\lambda}$.Given $h = 6.63 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$,and $\lambda = 0.021 	imes 10^{-9} \ m$.$\Delta E = \frac{(6.63 	imes 10^{-34} \ J \cdot s) 	imes (3 	imes 10^8 \ m/s)}{0.021 	imes 10^{-9} \ m}$.$\Delta E \approx 9.47 	imes 10^{-15} \ J$.To convert this energy into $keV$,divide by the charge of an electron $(1.6 	imes 10^{-19} \ C)$ and then by $1000$.$\Delta E_{keV} = \frac{9.47 	imes 10^{-15}}{1.6 	imes 10^{-19} 	imes 1000} \approx 59.2 \ keV$.Rounding to the nearest integer,the energy difference is $59 \ keV$.

The $K_\alpha$ $X$-ray emission line for tungsten occurs at $\lambda = 0.021 \ nm$. What is the energy difference between the $K$ and $L$ levels in this atom in $keV$?

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