Electrons with energy `80 keV` are incident on the tungsten target of an X - rays tube , k- shell electrons of tungsten have `72.5 keV` energy X- rays emitted by the tube contain only

Options

A. a continous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of `= 0.155 Å`
B. a continous X-ray spectrum (Bremsstrahlung) with all wavelengths
C. the characterstic X-ray spectrum of tungsten
D. a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of `= 0.155 Å` and the characteristic X-ray spectrum of tungsten.

Answer

D

Explanation

Minimum wavelength of continuous X-ray spectrum is given by
`lambda_("min")` (in Å)`=(12375)/("E(in eV)")`
Here, E= energy of incident electron (in eV)
= energy corresponding to minimum wavelength
`implies lambda_("min")` of X-ray `implies E=80 keV=80 xx10^(3) eV`
`:. lambda_("min")` (in Å) `=12375/(80xx10^(3))~~0.155`
Also the energy of the incident electrons `(80 keV)` is more than the ionization energy of the K-shell electrons (i.e. 72.5 keV). 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.