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(D) The shortest wavelength (cut-off wavelength) of $X$-rays produced is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400 \ 	ext{e

Vedclass · Accepted Answer

$0.025 \ nm$

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(D) The shortest wavelength (cut-off wavelength) of $X$-rays produced is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400 \ 	ext{eV} \cdot \mathring{A}}{V 	ext{ (in volts)}}$.Given the potential difference $V = 50000 \ V$.Substituting the values: $\lambda_{\min} = \frac{12400}{50000} \ \mathring{A} = 0.248 \ \mathring{A} \approx 0.0248 \ nm$.Rounding to the nearest provided option,we get $0.025 \ nm$.

An electron beam in an $X$-ray tube is accelerated through a potential difference of $50000 \ V$. These are then made to fall on a tungsten target. The shortest wavelength of the $X$-ray emitted by the tube is

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