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. The target is kept at a potential V higher than the cathode resulting in emission of continuous and characteristic X-rays. If the filament current 𝐼 is decreased to $\frac{I}{2}$ , the potential difference 𝑉 is increased to 2V, and the separation distance 𝑑 is reduced to $\frac{d}{2}$ , then

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the cut-off wavelength will reduce to half, and the wavelengths 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

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

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

answer is A, C.

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Detailed Solution

$\begin{array}{l} λ_{min} = \frac{hc}{eV} \\ \Rightarrow λ_{min} α \frac{1}{V} \\ \Rightarrow {(λ_{min})}_{new} = \frac{λ_{2}}{2} \\ Characteristic X-ray depends on target atomic number so it remain same \\ ∵ I = \frac{dN}{dt} \times \frac{hc}{λ} \\ ∵ \frac{dN}{dt} decreases \\ On decreasing filament current, number of electrons decrease . So intensity of X-ray decreases. \\ Hence I decreases \end{array}$