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(D) The wavelength of the $K_{\alpha}$ line is characteristic of the target material and is independent of the accelerating voltage $V$. Thus,$\lambda

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Becomes less than two times

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(D) The wavelength of the $K_{\alpha}$ line is characteristic of the target material and is independent of the accelerating voltage $V$. Thus,$\lambda_{K_{\alpha}}$ remains constant.The minimum wavelength of the continuous $X-ray$ spectrum is given by $\lambda_{\min} = \frac{hc}{eV}$.When the voltage $V$ is halved to $V' = V/2$,the new minimum wavelength becomes $\lambda'_{\min} = \frac{hc}{e(V/2)} = 2\lambda_{\min}$.The initial difference is $\Delta \lambda = \lambda_{K_{\alpha}} - \lambda_{\min}$.The new difference is $\Delta \lambda' = \lambda_{K_{\alpha}} - 2\lambda_{\min}$.Since $\lambda_{K_{\alpha}} > \lambda_{\min}$,we can write $\Delta \lambda' = \lambda_{K_{\alpha}} - 2\lambda_{\min} = (\lambda_{K_{\alpha}} - \lambda_{\min}) - \lambda_{\min} = \Delta \lambda - \lambda_{\min}$.Since $\lambda_{\min} > 0$,it follows that $\Delta \lambda' Also,$2\Delta \lambda = 2\lambda_{K_{\alpha}} - 2\lambda_{\min}$. Comparing $\Delta \lambda' = \lambda_{K_{\alpha}} - 2\lambda_{\min}$ with $2\Delta \lambda$,we see that $\Delta \lambda' < 2\Delta \lambda$. Therefore,the difference becomes less than two times the original difference.

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

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