JEE Advanced 2020 · Paper 2 · Physics Q10

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PhysicsMultiple correct+4 / -2Question 10 of 54

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

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 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

Answer and worked solution

Official answer: A, C

Cut-off wavelength. $\lambda_{min}=\dfrac{hc}{eV}$ , so doubling $V$ halves it. The distance $d$ does not matter.

Characteristic lines. They come from transitions in the target atoms, so their wavelengths are unchanged.

Intensity. Halving the filament current halves the number of electrons hitting the target, so the intensities of all X-rays decrease.

(A), (C)

Topics:X-rays

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JEE Advanced 2020 · Paper 2, Physics question 10