In X-ray tube when the accelerating voltage $V$ is halved, the difference between the wavelengths of $K_{\alpha }$ line and minimum wavelength of continuous X-ray spectrum

Since, $\Delta \lambda ={\lambda }_{K_{\alpha }}-{\lambda }_{\min }$

When $V$ is made half, ${\lambda }_{\min }$ becomes two times, however ${\lambda }_{K_a}$ remains the same.

$\Rightarrow \Delta {\lambda }^{\text{'}}={\lambda }_{K_a}-2{\lambda }_{\min }$

From (1), we have

$$\begin{gathered} {\lambda }_{\min }={\lambda }_{K_{\alpha }}-\Delta \lambda \\ \Rightarrow \Delta {\lambda }^{\text{'}}={\lambda }_{K_{\alpha }}-2{\lambda }_{K_{\alpha }}+2\Delta \lambda \\ \Rightarrow \Delta {\lambda }^{\text{'}}=2\Delta \lambda -{\lambda }_{K_{\alpha }} \\ \Rightarrow \Delta {\lambda }^{\text{'}}<2\Delta \lambda \end{gathered}$$

Hence, the correct answer is (D).