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(D) The minimum wavelength (cutoff wavelength) of $X$-rays produced by electrons accelerated through a potential difference $V$ is given by the Duane-

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$40$

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(D) The minimum wavelength (cutoff wavelength) of $X$-rays produced by electrons accelerated through a potential difference $V$ is given by the Duane-Hunt law:$\lambda_{\min} = \frac{hc}{eV}$Substituting the values of constants $h = 6.626 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$,and $e = 1.602 	imes 10^{-19} \ C$,the formula simplifies to:$\lambda_{\min} (	ext{in } \mathring{A}) \approx \frac{12400}{V (	ext{in } V)}$ or $\lambda_{\min} (	ext{in } \mathring{A}) \approx \frac{12.4}{V (	ext{in } kV)}$Given $\lambda_{\min} = 0.3094 \ \mathring{A}$,we have:$0.3094 = \frac{12400}{V}$$V = \frac{12400}{0.3094} \approx 40077 \ V = 40.077 \ kV$Rounding to the nearest given option,we get $V \approx 40 \ kV$.

In producing $X$-rays,a beam of electrons accelerated by a potential difference $V$ is made to strike a metal target. For what value of $V$ will the $X$-rays have the lowest wavelength of $0.3094 \ \mathring{A}$? (in $kV$)

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