Question

The energies for an electron in the K,L and M shells of the tungsten atom are -69,500 eV, -12.000 eV, and -2200 eV respectively. Calculate the wavelengths of the${\mathbf{K}}_{\mathbf{\alpha }}$ and${\mathbf{K}}_{\mathbf{\beta }}$ x-rays of tungsten.

Short answer

a) The wavelength of ${\mathrm{K}}_{\mathrm{\alpha }}$x-ray of tungsten is 21.6 pm.

b) The wavelength of ${\mathrm{K}}_{\mathrm{\beta }}$x-ray of tungsten is 18.4 pm.

Step-by-step solution

Define Moseley’s Law.

If the frequency of${\mathbf{K}}_{\mathbf{\alpha }}$ x-ray of the spectrum of an element of an atomic number Z is f, then by Moseley’s law frequency is written as:

$\mathbf{f}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{48}\mathbf{\times }{\mathbf{10}}^{\mathbf{15}}\mathbf{Hz}\mathbf{)}\mathbf{(}\mathbf{Z}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}$……………(1)

The energy of the photon is determined by the equation,

$\mathbf{E}\mathbf{=}\mathbf{hf}$.

The relation between the wavelength and frequency of the photon is,

$\mathbf{\lambda }\mathbf{=}\frac{\mathbf{c}}{\mathbf{f}}$.

Determine the wavelength of Kα x-rays of tungsten.

Given that, the energies for an electron in the K,L and M shells of the tungsten atom are $-69,500\mathrm{eV},-12,000\mathrm{eV}$and -2200 eV.

(a)

The difference between the k- shell and L-shell energies is the energy of the$K_{\alpha }$ x-ray:

$$\begin{gathered} {\mathrm{E}}_{{\mathrm{K}}_{\mathrm{\alpha }}}=\left(-12,000\mathrm{eV}\right)-\left(-69,500\mathrm{eV}\right) \\ =57500\mathrm{eV} \end{gathered}$$

So, the wavelength of$K_{\alpha }$ x-ray of tungsten is:

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{\alpha }}=\frac{\mathrm{hc}}{\mathrm{E}} \\ =\frac{\left(4.316\times {10}^{-15}\mathrm{eV}.\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}{57500\mathrm{eV}} \\ =2.16\times {10}^{-11}\mathrm{m}\left(\frac{1\mathrm{pm}}{{10}^{-12}\mathrm{m}}\right) \\ =21.6\mathrm{pm} \end{gathered}$$

Hence, the wavelength of$K_{\alpha }$ x-ray of tungsten is 21.6 pm.

Determine the wavelength of Kβ x-rays of tungsten.

(b)

The difference between the K-shell and M-shell energies is the energy of the$K_{\beta }$ x-ray:

$$\begin{gathered} {\mathrm{E}}_{{\mathrm{K}}_{\mathrm{\beta }}}=\left(-2200\mathrm{eV}\right)-\left(-69,500\mathrm{eV}\right) \\ =67300\mathrm{eV} \end{gathered}$$

So, the wavelength of${\mathrm{K}}_{\mathrm{\beta }}$ x-ray of tungsten is:

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{\beta }}=\frac{\mathrm{hc}}{\mathrm{E}} \\ =\frac{\left(4.316\times {10}^{-15}\mathrm{eV}.\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}{67300\mathrm{eV}} \\ =1.84\times {10}^{-11}\mathrm{m}\left(\frac{1\mathrm{pm}}{{10}^{-12}\mathrm{m}}\right) \\ =18.4\mathrm{pm} \end{gathered}$$

Hence, the wavelength of$K_{\beta }$ x-ray of tungsten is 18.4 pm.