The K_{alpha} X-ray emission line of tungsten | vedclass

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(C) The energy of the emitted $X$-ray photon corresponds to the energy difference between the two levels involved in the transition.For the $K_{\alpha

Vedclass · Accepted Answer

$59 \ keV$

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(C) The energy of the emitted $X$-ray photon corresponds to the energy difference between the two levels involved in the transition.For the $K_{\alpha}$ line,the transition occurs from the $L$ level to the $K$ level.The energy difference is given by $\Delta E = E_K - E_L = \frac{hc}{\lambda}$.Substituting the values: $h = 6.6 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$,and $\lambda = 0.021 	imes 10^{-9} \ m$.$\Delta E = \frac{(6.6 	imes 10^{-34}) 	imes (3 	imes 10^8)}{0.021 	imes 10^{-9}} \ J$.To convert this energy into electron-volts $(eV)$,divide by $1.6 	imes 10^{-19} \ J/eV$:$\Delta E = \frac{19.8 	imes 10^{-26}}{0.021 	imes 10^{-9} 	imes 1.6 	imes 10^{-19}} \ eV \approx 58928 \ eV \approx 59 \ keV$.

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

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