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 keV,11.3 keV, and 2.30 keV respectively. (a) What is the minimum value of the accelerating potential that will permit the production of the characteristic${\mathbf{k}}_{\mathbf{\alpha }}$ and${\mathbf{k}}_{\mathbf{\beta }}$ lines of tungsten? (b) For this same accelerating potential, what is ${\mathbf{\lambda }}_{\mathbf{min}}$? What are the (c) ${\mathbf{k}}_{\mathbf{\alpha }}$and (d)${\mathbf{k}}_{\mathbf{\beta }}$ wavelengths?

The atomic number of the target tungsten, Z=74

The energies for the K,L and M energy levels for tungsten are 69.5 keV, 11.3 keV, and 2.30 keV respectively.

Consider the known data below.

The Plank’s constant is,

$$\begin{gathered} \mathrm{h}=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s} \\ =\left(6.242\times {10}^{15}\times 6.63\times {10}^{-34}\right)\mathrm{keV}.\mathrm{s} \\ =41.384\times {10}^{-19}\mathrm{keV}.\mathrm{s} \end{gathered}$$

The speed of light is,

$$\begin{gathered} \mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s} \\ =\left(3\times {10}^8\times {10}^{12}\right)\mathrm{pm}/\mathrm{s} \\ =3\times {10}^{20}\mathrm{pm}/\mathrm{s} \end{gathered}$$

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

Use the concept of Planck's relation for the value of the minimum wavelength that permits the removal of an electron from the lowest energy level. Again, the line is produced only when an electron jumps from K to L energy level. Thus, the energy difference between these levels will provide the required energy. Similarly, the energy difference for the K line for the K to M energy jump can be calculated.

Formula:

The energy of the photon due to Planck’s relation,

$∆\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ….. (1)

Here,$∆\mathrm{E}$ is the energy of the photon,h is the Plank’s constant, and c is the speed of light.

An electron must be removed from the K-shell, so that an electron from a higher energy shell can drop. This requires energy of 69.5 keV.

Hence, the accelerating potential must be at least 69.5 keV that will permit the production of the characteristic lines.

Using the same accelerating potential from above part in equation (a), the minimum wavelength of the line can be given as follow.

$$\begin{gathered} {\mathrm{\lambda }}_{\min }=\frac{\mathrm{hc}}{∆\mathrm{E}} \\ =\frac{\left(41.384\times {10}^{-19}\mathrm{keV}.\mathrm{s}\right)\left(3\times {10}^{20}\mathrm{pm}/\mathrm{s}\right)}{69.5\mathrm{keV}} \\ =17.8\mathrm{pm} \end{gathered}$$

Hence, the value of the wavelength is 17.8 pm.

The energy of a photon associated with the$K_{\alpha }$ line can be given using the data as follows:

$$\begin{gathered} {\mathrm{E}}_{{\mathrm{K}}_{\mathrm{\alpha }}}=\left(69.5-11.3\right)\mathrm{keV} \\ =58.2\mathrm{keV} \end{gathered}$$

Now, using this energy value in equation (1), the wavelength of the${\mathrm{K}}_{\mathrm{\alpha }}$ line as follows:

$$\begin{gathered} {\mathrm{\lambda }}_{{\mathrm{K}}_{\mathrm{\alpha }}}=\frac{\mathrm{hc}}{{\mathrm{E}}_{{\mathrm{K}}_{\mathrm{\alpha }}}} \\ =\frac{\left(41.384\times {10}^{-19}\mathrm{keV}.\mathrm{s}\right)\left(3\times {10}^{20}\mathrm{pm}/\mathrm{s}\right)}{58.2\mathrm{keV}} \\ =21.3\mathrm{pm} \end{gathered}$$

Hence, the value of the wavelength is 21.3 pm.

The energy of a photon associated with the${\mathrm{K}}_{\mathrm{\beta }}$ line can be given using the data as follows:

$$\begin{gathered} {\mathrm{E}}_{{\mathrm{K}}_{\mathrm{\beta }}}=\left(69.5-2.30\right)\mathrm{keV} \\ =67.2\mathrm{keV} \end{gathered}$$

Now, using this energy value in equation (i), we can get the wavelength of the$K_{\beta }$ line as follows:

$$\begin{gathered} {\mathrm{\lambda }}_{{\mathrm{K}}_{\mathrm{\beta }}}=\frac{\mathrm{hc}}{{\mathrm{E}}_{{\mathrm{K}}_{\mathrm{\beta }}}} \\ =\frac{\left(41.384\times {10}^{-19}\mathrm{keV}.\mathrm{s}\right)\left(3\times {10}^{20}\mathrm{pm}/\mathrm{s}\right)}{67.2\mathrm{keV}} \\ =18.5\mathrm{pm} \end{gathered}$$

Hence, the value of the wavelength is 18.5 pm.