Question 3

The wavelength of the $K_\alpha$ line is given by Moseley's law: $\frac{1}{\lambda_{K_\alpha}} \propto (Z-1)^2$, so $\lambda_{K_\alpha} = \frac{C}{(Z-1)^2}$. The cut-off wavelength is $\lambda_{min} = \frac{hc}{eV}$. Since the same electron beam (same voltage $V$) is used, $\lambda_{min}$ is constant.

The ratio is $r = \frac{\lambda_{K_\alpha}}{\lambda_{min}} = \frac{C}{\lambda_{min}(Z-1)^2}$. For $Z = 46$, $r = 2$: $2 = \frac{C}{\lambda_{min}(46-1)^2} = \frac{C}{\lambda_{min}(45)^2} \implies \frac{C}{\lambda_{min}} = 2 \times 45^2$ For $Z = 41$: $r' = \frac{C}{\lambda_{min}(41-1)^2} = \frac{2 \times 45^2}{40^2} = 2 \times \left(\frac{9}{8}\right)^2 = 2 \times \frac{81}{64} = \frac{81}{32} = 2.53125 \approx 2.53$