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

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(C) The energy of an $X$-ray photon is given by the relation $E = \frac{hc}{\lambda}$.Given $\lambda = 0.021 \, nm = 0.21 \, \mathring{A}$.Using the a

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$59 \, keV$

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(C) The energy of an $X$-ray photon is given by the relation $E = \frac{hc}{\lambda}$.Given $\lambda = 0.021 \, nm = 0.21 \, \mathring{A}$.Using the approximation $E \approx \frac{12400 \, eV \cdot \mathring{A}}{\lambda (	ext{in } \mathring{A})}$,we get:$E = \frac{12400}{0.21} \, eV \approx 59047 \, eV$.Since the $K_{\alpha}$ line corresponds to the transition from the $L$-shell to the $K$-shell,the energy difference is $\Delta E = E_L - E_K = E_{photon} \approx 59 \, keV$.

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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