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(D) The minimum wavelength of the continuous $X$-ray spectrum is given by the formula $\lambda_{\min} = \frac{hc}{E}$.Using $hc \approx 12400 \ eV \cd

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$A$ continuous $X$-ray spectrum (Bremsstrahlung) with a minimum wavelength of $\approx 0.155 \ \mathring{A}$ and the characteristic $X$-ray spectrum of tungsten

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(D) The minimum wavelength of the continuous $X$-ray spectrum is given by the formula $\lambda_{\min} = \frac{hc}{E}$.Using $hc \approx 12400 \ eV \cdot \mathring{A}$ (or $12375 \ eV \cdot \mathring{A}$),we get $\lambda_{\min} = \frac{12375}{80 	imes 10^3} \approx 0.155 \ \mathring{A}$.Since the energy of the incident electrons $(80 \ keV)$ is greater than the ionization energy of the $K$-shell electrons $(72.5 \ keV)$,the incident electrons have sufficient energy to eject $K$-shell electrons from the tungsten atoms.When an electron is ejected from the $K$-shell,electrons from higher energy levels transition to fill the vacancy,resulting in the emission of characteristic $X$-rays.Therefore,the emitted $X$-rays consist of both a continuous spectrum (Bremsstrahlung) and the characteristic $X$-ray spectrum of tungsten.

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

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