JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Modern Physics

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

2 V

, and the separation distance

d

is reduced to

\frac{d}{2}

, then

(A)the cut-off wavelength will reduce to half, and the wavelengths of the characteristic X-rays will remain the same✓ Correct
(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✓ Correct
(D)the cut-off wavelength will become two times larger, and the intensity of all the X-rays will decrease

Answer: (A), (C)

Solution

Cut-off wavelength:

λ_{min} = \frac{h c}{e V}

Since

λ_{min} \propto \frac{1}{V}

, when V is doubled:

(λ_{min})_{new} = \frac{λ _{min}}{2}

✓

Characteristic X-ray wavelengths depend only on the target material, not on the accelerating voltage, so they remain the same. ✓ (Option A)

Intensity:

I = \frac{d N}{d t} \times \frac{h c}{λ}

Since filament current is halved,

\frac{d N}{d t}

(rate of electrons) decreases, so intensity of all X-rays decreases. ✓ (Option C)

Note: The distance d does not affect the wavelengths or intensities in the standard X-ray tube model.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · One or more correct options

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Modern Physics

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