JEE Advance - Physics (2020 - Paper 2 Offline - No. 10)
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 $${1 \over 2}$$, the potential difference V is increased to 2V, and the separation distance d is reduced to $${d \over 2}$$, 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
Explanation
$$\lambda \propto {1 \over V}$$
So, when V double, then cut-off wavelength become half. Characteristic does not depend on voltage. When I is halved, then intensity will decrease.
So, when V double, then cut-off wavelength become half. Characteristic does not depend on voltage. When I is halved, then intensity will decrease.
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