17.4.3 Bremsstrahlung Radiation

When we were discussing the problem of the Rutherford’s atom, we mentioned that acceleration of electric charges must be accompanied by EM radiation. As a matter of fact, when we push a high frequency AC current in copper wires, radio waves are emitted. This is how antennas work! So can we produce x-rays by oscillating an electron vigorously?

Usually not. The magnitude of acceleration needed for a charge to emit x-rays is incredibly large. Even the most powerful AC supply in this world cannot provide that kind of acceleration needed. But the nuclei of a tungsten atom can.

In the x-ray tube, some of the filament electrons can get really close to the tungsten nuclei[1]. The nucleus of a tungsten atom is a +74e of (positive) charge squeezed into a tiny volume. It is almost like the proverbial point charge whose field strength approaches infinity at close distance. As a result, these filament electrons are deflected sharply as they are “sucked” towards the nucleus. The acceleration they undergo is so strong that x-ray photons are emitted in the process.

The bremsstrahlung photon energy depends on the degree of interaction between the electron and the nucleus. The closer the electron swings past the nucleus, the more KE it loses, and the more energetic the x-ray photon produced. Since the filament electron can lose any fraction of its energy during an interaction with the nucleus, it can lose anything between 0 keV and 100 keV of energy (assuming our X-ray tube has an accelerating voltage of 100 kV). This implies that a bremsstrahlung photon can have any energy between 0 keV and 100 keV.

This is the reason why the bremsstrahlung radiation has a continuous spectrum. Furthermore, the cut-off frequency (and cut-off wavelength) can be calculated easily by equating the x-ray photon energy to the KE of the filament electron.

$\displaystyle h{{f}_{{\max }}}=\frac{{hc}}{{{{\lambda }_{{\min }}}}}=e{{V}_{A}}$

Did you notice that the continuous spectrum is depicted with a minimum frequency (besides the maximum cut-off frequency), which seems to suggest that low frequency photons are not emitted. Actually, the bremsstrahlung process produces more low frequency photons than high frequency photons. Just that in practice, the longer wavelength radiations are naturally absorbed and filtered away by the walls of the x-ray tube.

Braking Radiation

As the filament electron swings around the tungsten nucleus, it actually speeds up when it is approaching the nucleus, but slows down on its way out. But overall, the electron has less KE after its interaction with the nucleus, since some KE has been converted into X-ray photons. This is the reason why bremsstrahlung radiation is also called the braking radiation. This is analogues to a ball rolling through a pit with a rough surface; The ball accelerates while rolling down and decelerates while rolling up. But overall, the ball exits the hole with less KE than it entered with, since some KE has been converted into heat energy.

Example

The K and L shell ionization energies of copper are 8979 eV and 951 eV respectively. For a x-ray tube that uses accelerating voltage V_A of 12 kV, determine

the wavelength of the K_a line
minimum cut-off wavelength.

Solution

$\displaystyle \displaystyle \begin{aligned}\frac{{hc}}{\lambda }&=\Delta E\\\frac{{(6.63\times {{{10}}^{{-34}}})(3.00\times {{{10}}^{8}})}}{\lambda }&=(8979-951)(1.60\times {{10}^{{-19}}})\\\lambda &=1.55\times {{10}^{{-10}}}\text{ m}\end{aligned}$

$\displaystyle \displaystyle \begin{aligned}\frac{{hc}}{{{{\lambda }_{{\min }}}}}&=e{{V}_{A}}\\\frac{{(6.63\times {{{10}}^{{-34}}})(3.00\times {{{10}}^{8}})}}{\lambda }&=(12000)(1.60\times {{10}^{{-19}}})\\\lambda &=1.04\times {{10}^{{-10}}}\text{ m}\end{aligned}$