The wavelength of {K_alpha } X-rays produced | vedclass

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(C) The wavelength of $X-$ray lines is given by Moseley's law,which is a modification of the Rydberg formula: $\frac{1}{\lambda } = R(Z - 1)^2 \left(

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$40$

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(C) The wavelength of $X-$ray lines is given by Moseley's law,which is a modification of the Rydberg formula: $\frac{1}{\lambda } = R(Z - 1)^2 \left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} ight)$.For the ${K_\alpha }$ line,${n_1} = 1$ and ${n_2} = 2$.Substituting these values: $\frac{1}{\lambda } = R(Z - 1)^2 \left( {\frac{1}{1^2} - \frac{1}{2^2}} ight) = R(Z - 1)^2 \left( {\frac{3}{4}} ight)$.Rearranging for $Z$: $(Z - 1)^2 = \frac{4}{3R\lambda }$.Given $R \approx 1.097 	imes 10^7 \ m^{-1}$ and $\lambda = 0.76 	imes 10^{-10} \ m$:$(Z - 1)^2 = \frac{4}{3 	imes (1.097 	imes 10^7) 	imes (0.76 	imes 10^{-10})} \approx \frac{4}{0.0025} \approx 1600$.Taking the square root: $Z - 1 = \sqrt{1600} = 40$.Therefore,$Z = 41 \approx 40$ (depending on the precision of $R$ and $\lambda$ used,$40$ is the closest integer).

The wavelength of ${K_\alpha }$ $X-$rays produced by an $X-$ray tube is $0.76 \ \mathring A$. The atomic number of the anode material of the tube is:

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