The ratio of the energy of the photon emitted | vedclass

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(B) According to Moseley's law,the energy of a photon emitted during an electronic transition in an atom is given by $\Delta E = Rch(Z-1)^2 \left( \fr

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less than $1$

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(B) According to Moseley's law,the energy of a photon emitted during an electronic transition in an atom is given by $\Delta E = Rch(Z-1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the $K_{\alpha}$ line,the transition is from $n_2 = 2$ to $n_1 = 1$. Thus,$\Delta E_{\alpha} \propto \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \left( 1 - \frac{1}{4} \right) = \frac{3}{4} = 0.75$.For the $K_{\beta}$ line,the transition is from $n_2 = 3$ to $n_1 = 1$. Thus,$\Delta E_{\beta} \propto \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = \left( 1 - \frac{1}{9} \right) = \frac{8}{9} \approx 0.889$.Since $\Delta E_{\alpha} < \Delta E_{\beta}$,the ratio $\frac{\Delta E_{\alpha}}{\Delta E_{\beta}}$ is less than $1$.

The ratio of the energy of the photon emitted by the $K_{\alpha}$ line to that of the $K_{\beta}$ line is

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