JEE Challenger
More from Dual Nature of Radiation and Matter
PhyMedAdv 2020 • Paper 2

Cut Off Wavelength and Intensity Changes in X Ray Tube Setup

Consider an X-ray tube where electrons are emitted from a cathode filament carrying a current II and hit an anode target situated at a distance dd from the cathode. The target is held at a potential VV greater than that of the cathode, producing continuous and characteristic X-rays. If the filament current II is reduced to I2\frac{I}{2}, the potential difference VV is doubled to 2V2V, and the separation dd is halved to d2\frac{d}{2}, then

Options

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

D

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

Topics & Concepts

Dual Nature of Radiation and MatterX-Rays

Step-by-Step Solution

To determine the correct statement(s), we analyze the effects of changing the parameters (filament current II, potential difference VV, and separation distance dd) on the emitted X-ray spectrum:

1. Cut-off Wavelength (λmin\lambda_{\text{min}})

The continuous X-ray spectrum is produced by bremsstrahlung (braking radiation). The maximum photon energy corresponds to an electron transferring all of its kinetic energy eVeV into a single photon. By the Duane–Hunt law, the cut-off (minimum) wavelength is given by: λmin=hceV\lambda_{\text{min}} = \frac{hc}{eV}

When the potential difference is doubled from VV to V′=2VV' = 2V: λmin′=hce(2V)=12(hceV)=λmin2\lambda_{\text{min}}' = \frac{hc}{e(2V)} = \frac{1}{2} \left(\frac{hc}{eV}\right) = \frac{\lambda_{\text{min}}}{2}

Thus, the cut-off wavelength reduces to half of its original value.

2. Wavelengths of Characteristic X-rays

Characteristic X-rays are emitted when an incoming high-energy electron knocks out an inner-shell electron of a target atom, causing an electron from a higher energy level to transition down to fill the vacancy.

The wavelengths of characteristic X-rays are governed by Moseley's law and depend exclusively on:

  • The atomic number ZZ of the target material.
  • The energy levels of the specific atomic transitions (e.g., Kα,KβK_\alpha, K_\beta).

Since the target material is not changed, the energy level differences remain constant. Therefore, the wavelengths of the characteristic X-rays remain the same, independent of II, VV, or dd.

3. Intensity of X-rays

  • The filament current II controls the thermionic emission rate of electrons, which determines the tube current ii. Reducing II decreases the number of electrons hitting the target per unit time.
  • However, the total intensity of continuous X-rays is proportional to iV2i V^2. Halving the current (i→i/2i \to i/2) while doubling the voltage (V→2VV \to 2V) results in an intensity proportional to: Intensity∝(i2)(2V)2=2iV2\text{Intensity} \propto \left(\frac{i}{2}\right)(2V)^2 = 2 i V^2 This causes the overall intensity of continuous X-rays to increase rather than decrease. Thus, options (C) and (D) are incorrect.

Conclusion

  • The cut-off wavelength reduces to half.
  • The wavelengths of the characteristic X-rays remain unchanged.

Therefore, the correct option is A.

PreviousNext