Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

In an X - ray tube, electrons emitted from a filament (cathode) carrying current I hit a target 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 characteristics X rays . if the filament current I is decreased to I/2, the potential difference V is increased to 2V, and the wavelengths of the characteristics X rays will be ...

solution : cut off wavelength can be given by, λ_min = hc/eV

where h is plank's constant, c is speed of light , e is charge on electron and V is potential difference.

here λ_min ∝ 1/V

so if the potential difference is increased to 2V , cut off wavelength becomes half.

therefore option (A) is correct choice.

because intensity is given by, I = (dN)/dt × hc/λ

here dN/dt is the no of electrons emitted from filament.

on decreasing filament current, number of electrons decreases so intensity of x - rays must be decreases.

Therefore option (C) is also correct choice.

Therefore options (A) and (C) are correct choices.