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Q.

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 I2, the potential difference 𝑉 is increased to 2V, and the separation distance 𝑑 is reduced to d2 , then

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a

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

b

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

c

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

d

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

answer is A.

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Detailed Solution

λmin=hceV⇒λminα1V⇒λminnew =λ22Characteristic X-ray depends on target atomic number so it remain same ∵I=dNdt×hcλ∵dNdt decreases On decreasing filament current, number of electrons decrease .  So intensity of X-ray decreases. Hence I decreases
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