## Related questions with answers

A tungsten (Z = 74) target is bombarded by electrons in an x-ray tube. The K, L, and M energy levels for tungsten 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

${\lambda}_{min}$

? What are the (c)

$K_{\alpha}$

and (d)

$K_{\beta}$

wavelengths?

Solution

Verified$\textbf{(a)}$ To produce a characteristic $K_\alpha$ and $K_\beta$ lines of tungsten, an electron must be removed from the $K$-shell the energy required to remove an electron form this shell is just the energy of this shell. The $K$ energy level for tungsten have an energy of 69.5 keV, thus the minimum potential difference that can be used to remove an electron from $K$ level is:

$\boxed{V=69.5 \mathrm{~kV}}$

$\textbf{(b)}$ The minimum wavelength is given by:

$\lambda_\text{min}=\dfrac{hc}{E}$

where $E$ is the energy of the $K$ level, so:

$\begin{align*}\lambda_\text{min}&=\dfrac{(6.626 \times 10^{-34} \mathrm{~J\cdot s})(2.998 \times 10^8 \mathrm{~m/s})}{69.5 \times 10^3 \times 1.602 \times 10^{-19} \mathrm{~J}} \\ &=1.78 \times 10^{-11} \mathrm{~m}\\ &=17.8 \mathrm{~pm} \end{align*}$

$\boxed{\lambda_\text{min}=17.8 \mathrm{~pm}}$

$\textbf{(c)}$ According to figure 40-15, the energy of a photon associated with the $K_\alpha$ line is given by:

$\begin{align*} E_{K_\alpha}&=E_K-E_L\\ &=69.5 \mathrm{~keV}-11.3\mathrm{~keV}\\ &=58.2\mathrm{~keV} \end{align*}$

and the wavelength that corresponds this energy is:

$\begin{align*}\lambda&=\dfrac{(6.626 \times 10^{-34} \mathrm{~J\cdot s})(2.998 \times 10^8 \mathrm{~m/s})}{58.2 \times 10^3 \times 1.602 \times 10^{-19} \mathrm{~J}} \\ &=2.13\times 10^{-11} \mathrm{~m}\\ &=21.3\mathrm{~pm} \end{align*}$

$\boxed{\lambda=21.3\mathrm{~pm}}$

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