X-rays are produced by electrons accelerated through 24.8 kV onto a metal target. Use hc=12400 eV AA and Moseley's law √(nu)propto(Z-1) for the Kα…

(A) The shortest wavelength corresponds to the full electron energy: $\lambda_{\min}=\dfrac{12400}{24800}=0.50\ \text{\AA}$: true. (B) $\lambda_{\min}\propto1/V$, so doubling the voltage halves it to $0.25\ \text{\AA}$: true. (C) The characteristic lines depend only on the target's atomic energy levels, not on the accelerating voltage (provided it is above the excitation threshold), so $K_\alpha$ does not shift: false. (D) From Moseley's law $\lambda\propto1/(Z-1)^2$, so $\dfrac{\lambda_{21}}{\lambda_{41}}=\dfrac{40^2}{20^2}=4$: true. Only the continuous spectrum depends on the tube voltage; the characteristic lines are fingerprints of the target atoms, which is why the minimum wavelength can be tuned by the voltage but the $K_\alpha$ line cannot.