Let {lambda _alpha },{lambda _beta } and {lam | vedclass

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(C) According to the energy level diagram of $X-$ ray spectra,the energy of a photon emitted during an electronic transition is given by $\Delta E = \

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$\frac{1}{{{\lambda _\beta }}} = \frac{1}{{{\lambda _\alpha }}} + \frac{1}{{{{\lambda '}_\alpha }}}$

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(C) According to the energy level diagram of $X-$ ray spectra,the energy of a photon emitted during an electronic transition is given by $\Delta E = \frac{hc}{\lambda}$,which implies $\lambda \propto \frac{1}{\Delta E}$.For $X-$ ray transitions:$1$. The $K_\beta$ line corresponds to a transition from the $M$ shell to the $K$ shell.$2$. The $K_\alpha$ line corresponds to a transition from the $L$ shell to the $K$ shell.$3$. The $L_\alpha$ line corresponds to a transition from the $M$ shell to the $L$ shell.From the energy conservation principle,the energy of the $K_\beta$ transition is the sum of the energies of the $K_\alpha$ and $L_\alpha$ transitions:$(\Delta E)_{K_\beta} = (\Delta E)_{K_\alpha} + (\Delta E)_{L_\alpha}$Substituting $\Delta E = \frac{hc}{\lambda}$ into the equation:$\frac{hc}{\lambda_\beta} = \frac{hc}{\lambda_\alpha} + \frac{hc}{\lambda'_\alpha}$Dividing both sides by $hc$,we get:$\frac{1}{\lambda_\beta} = \frac{1}{\lambda_\alpha} + \frac{1}{\lambda'_\alpha}$

Let ${\lambda _\alpha }$,${\lambda _\beta }$ and ${\lambda '_\alpha }$ denote the wavelengths of the $X-$ rays of the ${K_\alpha }, {K_\beta }$ and ${L_\alpha }$ lines in the characteristic $X-$ rays for a metal.

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