The minimum wavelength of the X-rays produced | vedclass

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(C) The minimum wavelength (cut-off wavelength) of $X$-rays is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400}{V} 	ext{ Å}$.From

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$\lambda / 2$

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(C) The minimum wavelength (cut-off wavelength) of $X$-rays is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400}{V} 	ext{ Å}$.From this relation, we can see that $\lambda_{\min} \propto \frac{1}{V}$.Given that the initial potential is $V$ and the initial wavelength is $\lambda$, we have $\lambda = \frac{k}{V}$ (where $k$ is a constant).When the potential is changed to $V' = 2V$, the new minimum wavelength $\lambda'$ will be:$\lambda' = \frac{k}{V'} = \frac{k}{2V} = \frac{1}{2} \left( \frac{k}{V} ight) = \frac{\lambda}{2}$.

The minimum wavelength of the $X$-rays produced at an accelerating potential $V$ is $\lambda$. If the accelerating potential is changed to $2V$, then the minimum wavelength would become:

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