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(D) The minimum wavelength (cutoff wavelength) of an $X-ray$ tube is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.Substituting the values $h

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$5$

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(D) The minimum wavelength (cutoff wavelength) of an $X-ray$ tube is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.Substituting the values $h = 6.63 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$,and $e = 1.6 	imes 10^{-19} \ C$,we get the simplified relation: $\lambda_{\min} (	ext{in } \mathring{A}) \approx \frac{12400}{V (	ext{in volts})}$.Given $\lambda_{\min} = 2.5 	imes 10^{-10} \ m = 2.5 \ \mathring{A}$.Substituting this into the formula: $2.5 = \frac{12400}{V}$.$V = \frac{12400}{2.5} = 4960 \ V$.$V \approx 5000 \ V = 5 \ kV$.

If the minimum wavelength obtained in an $X-ray$ tube is $2.5 \times 10^{-10} \ m$,the operating potential of the tube will be.....$kV$.

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