The K, L and M energy levels of platinum lie | vedclass

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(D) The energy of a photon emitted during an electronic transition is given by $E = \frac{hc}{\lambda} = E_{initial} - E_{final}$.For the $K_{\alpha}$

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$\frac{25}{22}$

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(D) The energy of a photon emitted during an electronic transition is given by $E = \frac{hc}{\lambda} = E_{initial} - E_{final}$.For the $K_{\alpha}$ line,the transition is from the $L$ shell to the $K$ shell:$E_{K_{\alpha}} = \frac{hc}{\lambda_{K_{\alpha}}} = E_K - E_L = 78 \ keV - 12 \ keV = 66 \ keV$.For the $K_{\beta}$ line,the transition is from the $M$ shell to the $K$ shell:$E_{K_{\beta}} = \frac{hc}{\lambda_{K_{\beta}}} = E_K - E_M = 78 \ keV - 3 \ keV = 75 \ keV$.We need the ratio of the wavelength of the $K_{\alpha}$ line to that of the $K_{\beta}$ line:$\frac{\lambda_{K_{\alpha}}}{\lambda_{K_{\beta}}} = \frac{E_{K_{\beta}}}{E_{K_{\alpha}}} = \frac{75}{66} = \frac{25}{22}$.

The $K, L$ and $M$ energy levels of platinum lie roughly at $78, 12$ and $3 \ keV$ respectively. The ratio of the wavelength of the $K_{\alpha}$ line to that of the $K_{\beta}$ line in the $X$-ray spectrum is:

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