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

Option: 1

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


 


Option: 2

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


Option: 3

the characteristic X-ray spectrum of tungsten
 


Option: 4

a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of \mathrm{\approx 0.155 \AA} and the characteristic X-ray spectrum of tungsten.


Answers (1)

best_answer

Minimum wavelength of continuous X-ray spectrum is given by

\mathrm{ \lambda_{\min }(\text { in } \AA)=\frac{12375}{E(\text { in eV })} }

Here, \mathrm{ E= } energy of incident electrons (in \mathrm{ eV })

= energy corresponding to minimum wavelength

\mathrm{\lambda_{\min } \text { of X-rays } }

\mathrm{ E=80 \mathrm{keV} }

\mathrm{ =80 \times 10^3 \mathrm{eV} }

\mathrm{ \therefore \lambda_{\min }(\text { in } \AA)=\frac{12375}{80 \times 10^3} }

\mathrm{ \approx 0.155}

Also the energy of the incident electrons (80 \mathrm{keV}) is more than the ionization energy of the K-shell electrons  \mathrm{(i.e., 72.5 \mathrm{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 of L-shells.

Posted by

rishi.raj

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