The $K_{\alpha }$line in copper is a very common one to use in X-ray crystallography. To produce it, electrons are accelerated through a potential difference and smashed into a copper target. Section 7.8 gives the energies in a hydrogen like atom as${\mathit{Z}}^{\mathbf{2}}\left(-13.6\text{eV}/n^2\right)$ . Making the reasonable approximation that an$\mathit{n}\mathbf{=}\mathbf{1}$ electron in copper orbits the nucleus and half of its fellow $\mathit{n}\mathbf{=}\mathbf{1}$electron, being unaffected by the roughly spherical cloud of other electrons around it. Estimate the minimum accelerating potential needed to make a hole in copper'sKshell.

The given data is$K_{\alpha }$ line in copper is used in x-ray.

The energyEof the emitted ray is inversely proportional to the wavelength$\lambda$:

$E~\frac{1}{\lambda }$

Since a copper nucleus has 29 protons, its charge will be $+29$.

However, when the K-shell electron is ejected it sees half of the other K-shell electron screening the nucleus.

That will thus make the effective charge that the ejected electron sees as $+28.5$.

That is used in equation$E=Z^2\left(-\frac{13.6\text{eV}}{n^2}\right)$ for Z, along with 1 for (since it's the K-shell):

$$\begin{gathered} E=Z^2\left(-\frac{13.6\text{eV}}{n^2}\right) \\ E=(28.5{)}^2\left[-\frac{13.6\text{eV}}{{\left(1\right)}^2}\right] \\ E=-11046.6\text{eV} \end{gathered}$$

In order for the electron to be ejected, the sum of that and the kinetic energy of the accelerated electron have to be at least zero.

$$\begin{gathered} 0<KE+E \\ 0<q\Delta V-11046.6\text{eV} \\ q\Delta V-11046.6\text{eV} \end{gathered}$$

And since it's an electron that was accelerated, the charge q will be the elementary unit of charge q

$$\begin{gathered} 11046.6\text{eV}<q\Delta V \\ 11046.6\text{eV}<\left(e\right)\Delta V \\ 11046.6\text{eV}<\Delta V \end{gathered}$$

The electrons were able to drop out due to the definition of one $\text{VV}$ being the energy that an electron gains by passing through a potential difference of one Volt.

So, the potential difference that the X-ray electron would have to be accelerated through is shown below.

$$\begin{gathered} \Delta V=11,000\text{V} \\ \Delta V=11,000\text{V}\left(\frac{1\text{kV}}{1000\text{V}}\right) \\ \Delta V=11\text{kV} \end{gathered}$$