Question

Using a beam of electrons accelerated in an X-ray tube, we wish to knock an electron out of the shell of given element in a target. Section \(7.8\) gives the energies in a hydrogen like atom as . ${\mathit{Z}}^{\mathbf{2}}\left(-13.6\text{eV}/n^2\right)$Assume that for fairly high Z , aK-shell electron can be treated as orbiting the nucleus alone.

(a) A typical accelerating potential in an X-ray tube is$\mathbf{50}\mathbf{\text{kV}}$ . In roughly how high aZcould a hole in the K -shell be produced?

(b) Could a hole be produced in elements of higher Z?

Short answer

(a) The resultant answer is $Z=61$

(b) Yes, for higher.

Step-by-step solution

Given data

Accelerated potential, $\Delta V=50,000\text{V}$

Concept of Determinant

The energy of electron in${\text{n}}^{\text{th}}$-orbit:

The energy KE which an object of charge q gains by passing through the potential difference $\Delta V$is$KE=q\Delta V$

Determine the energy

(a)

If the accelerating potential is $V=50\text{kV}$, then the energy of the accelerated electron would be$E=50\text{keV}$ .

This electron can 'knock out' an electron in the shell with an energy of $E_1=-50\text{keV}$.

By Eq. (1), find the atomic number Z as (note that $n=1$ for the K shell).

$$\begin{gathered} Z=\sqrt{\frac{50·{10}^3\text{eV}}{13.6\text{eV}}} \\ Z\approx 60\backslash \mathrm{hfill} \end{gathered}$$

Determine the energy

(b)

If we were to strike an atom with higher atomic number than the one, we've calculated z=60 , we could make a hole but the level must be higher than $n=1$(the K shell).

If n was not larger, then the absolute value of the orbiting electron would be greater (see Eq. (1)) and the value of the accelerating potential would not be great enough.