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(D) The energy of the electron is given by $E = eV$,where $e$ is the charge of the electron and $V$ is the accelerating potential difference.This ener

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$40$

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(D) The energy of the electron is given by $E = eV$,where $e$ is the charge of the electron and $V$ is the accelerating potential difference.This energy is converted into the energy of the emitted $X$-ray photon,$E = \frac{hc}{\lambda}$.Equating the two,we get $eV = \frac{hc}{\lambda}$,which implies $\lambda = \frac{hc}{eV}$.Since $h$,$c$,and $e$ are constants,the minimum wavelength $\lambda_{min}$ is inversely proportional to the accelerating potential $V$ $(\lambda_{min} \propto \frac{1}{V})$.Therefore,to obtain the shortest wavelength,the potential difference $V$ must be the maximum among the given options.Comparing the options $10 \ kV, 20 \ kV, 30 \ kV,$ and $40 \ kV$,the maximum value is $40 \ kV$.

An electron beam is accelerated by a potential difference of $V$ and strikes a metal target to produce $X$-rays. For which of the following values of $V$ (in $kV$) will the $X$-ray wavelength be the shortest?

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