Question 40.43: A tungsten (Z = 74) target is bombarded by electrons in an x......
A tungsten (Z = 74) target is bombarded by electrons in an x-ray tube. The K, L, and M energy levels for tungsten (compare Fig. 40-15) have the energies 69.5, 11.3, and 2.30 keV, respectively. (a) What is the minimum value of the accelerating potential that will permit the production of the characteristic K_\alpha and K_\beta lines of tungsten? (b) For this same accelerating potential, what is λ_{min}? What are the (c) K_\alpha and (d) K_\beta wavelengths?
(a) An electron must be removed from the K-shell, so that an electron from a higher energy shell can drop. This requires an energy of 69.5 keV. The accelerating potential must be at least 69.5 kV.
(b) After it is accelerated, the kinetic energy of the bombarding electron is 69.5 keV. The energy of a photon associated with the minimum wavelength is 69.5 keV, so its wavelength is
\lambda_{\min }=\frac{1240\, eV \cdot nm }{69.5 \times 10^3 \,eV }=1.78 \times 10^{-2}\, nm =17.8 \,pm .
(c) The energy of a photon associated with the K_\alpha line is 69.5 keV – 11.3 keV = 58.2 keV and its wavelength is
\lambda_{ K \alpha}=(1240 \,eV \cdot nm ) /\left(58.2 \times 10^3 \,eV \right)=2.13 \times 10^{-2}\, nm =21.3\,pm .
(d) The energy of a photon associated with the K_\beta line is
E=69.5\, keV -2.30 \,keV =67.2 \,keV
and its wavelength is, using hc = 1240 eV·nm,
\lambda_{ K \beta}=h c / E=(1240 \,eV \cdot nm ) /\left(67.2 \times 10^3 \,eV \right)=1.85 \times 10^{-2}\, nm =18.5 \,pm .