The potential difference applied to an X-ray | vedclass

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(D) The minimum wavelength (cutoff wavelength) of $X$-rays produced in an $X$-ray tube is given by the formula: ${\lambda _{\min }} = \frac{{hc}}{{eV}

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The minimum wavelength decreases

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(D) The minimum wavelength (cutoff wavelength) of $X$-rays produced in an $X$-ray tube is given by the formula: ${\lambda _{\min }} = \frac{{hc}}{{eV}}$ where $h$ is Planck's constant,$c$ is the speed of light,$e$ is the charge of an electron,and $V$ is the accelerating potential difference. From the relation ${\lambda _{\min }} \propto \frac{1}{V}$,it is clear that the minimum wavelength is inversely proportional to the applied potential difference. Therefore,when the potential difference $V$ is increased,the minimum wavelength ${\lambda _{\min }}$ decreases. Thus,the correct option is $(d)$.

The potential difference applied to an $X$-ray tube is increased. As a result,in the emitted radiation

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