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(C) The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $K = eV$,where $e$ is the charge of the electr

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$\frac{1}{V}$

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(C) The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $K = eV$,where $e$ is the charge of the electron.When this electron strikes the target,the maximum energy of the emitted $X$-ray photon is equal to the kinetic energy of the electron,i.e.,$E_{\max} = h
u_{\max} = eV$.Since the frequency $
u = \frac{c}{\lambda}$,the minimum wavelength $\lambda_{\min}$ corresponds to the maximum energy:$\lambda_{\min} = \frac{hc}{E_{\max}} = \frac{hc}{eV}$.Since $h$,$c$,and $e$ are constants,we have $\lambda_{\min} \propto \frac{1}{V}$.Therefore,the correct option is $C$.

The minimum wavelength of $X$-rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to:

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