In an X-ray tube, electrons emitted from a filament (cathode) carrying current $I$ hit a target(anode) at a distance $d$ from the cathode resulting in emission of continuous and characteristic X-rays. If the filament current $I$ is decreased to $I / 2$ , the potential difference $V$ is increased to $2 V$ , and the separation distance $d$ is reduced to $d / 2$ , then :

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The cut-off wavelength will become two times larger, and the intensity of all the X-rays will decrease

The cut-off wavelength will reduce to half, and the wavelengths of the characteristic X-rays will remain the same

The cut-off wavelength as well as the wavelength of the characteristic X-rays will remain the same

The cut-off wavelength will reduce to half, and the intensities of all the X-rays will decrease

answer is A, C.

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Detailed Solution

Cut-off wavelength of X-rays is calculated as
$λ_{\min} = \frac{h c}{e V}$
$\Rightarrow λ_{\min} α \frac{h c}{e V}$
As potential difference across the tube in increased to 2V, the new cut-off wavelength becomes
${(λ_{\min})}_{n e w} = \frac{λ_{2}}{2}$
Wavelength of characteristic X-rays depend only upon the material of target anode so these will remain same hence option (A) is correct.
As filament current is decreased, number of electrons producing X-rays will decrease hence X-rays intensity will decrease. Hence option (C) is correct.