Modern Physics 4 Question 17

17. Electrons with energy 80keV are incident on the tungsten target of an X-ray tube. K-shell electrons of tungsten have 72.5keV energy. X-rays emitted by the tube contain only

(2000, 2M)

(a) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of 0.155\AA

(b) a continuous X-ray spectrum (Bremsstrahlung) with all wavelengths

(c) the characteristic X-ray spectrum of tungsten

(d) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of 0.155\AA and the characteristic X-ray spectrum of tungsten

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Solution:

  1. Minimum wavelength of continuous X-ray spectrum is given by λmin( in \AA)=12375E( in eV)

Here E= energy of incident electrons (in eV )

= energy corresponding to minimum wavelength λmin  of X-rays.

E=80keV=80×103eV

λmin ( in \AA)=1237580×1030.155

Also the energy of the incident electrons (80keV) is more than the ionization energy of the K-shell electrons (i.e., 72.5keV ). Therefore, characteristic X-ray spectrum will also be obtained because energy of incident electron is high enough to knock out the electron from K or L-shells.



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