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$\Delta \lambda$ is the difference between the wavelength of $k_\alpha$ line and the minimum wavelength of the continuous X-ray spectrum when the X-ray tube is operated at a voltage $V$. If the operating voltage is changed to $V / 3$, then the above difference is $\Delta \lambda^{\prime}$. Then
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$\Delta \lambda^{\prime}=3 \Delta \lambda$
$\begin{aligned} & \text { We know } \mathrm{eV}=\frac{h c}{\lambda} \Rightarrow \frac{V}{V^{\prime}}=\frac{\Delta \lambda^{\prime}}{\Delta \lambda} \\ & \Rightarrow \quad \frac{V}{V / 3}=\frac{\Delta \lambda^{\prime}}{\Delta \lambda} \\ & \Rightarrow \quad \Delta \lambda^{\prime}=3(\Delta \lambda)\end{aligned}$
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