An electron that is braked
An electron that strikes the target does not meet a solid wall. It enters a crowd of atoms, and it passes close to their nuclei. Each nucleus pulls on it with a strong electric force. The pull bends the electron's path and changes its speed. So near a nucleus the electron is accelerated, often very sharply.
An accelerated charge gives out electromagnetic radiation. This is a basic fact of electromagnetism, and the chapter on electromagnetic waves returns to it. Here the electron slows down, so the radiation takes away some of its kinetic energy. The radiation comes out as a single photon. It is called bremsstrahlung, German for braking radiation.
Energy is conserved in each braking, as the readouts of Fig 2.1 show. The nucleus is thousands of times heavier than the electron, so it takes almost no kinetic energy itself. The photon therefore carries exactly the kinetic energy that the electron lost. In the close pass, the electron goes from \(30\) to \(12\,\text{keV}\), and the photon carries \(18\,\text{keV}\).
When an electron is braked by a nucleus and gives out one photon, the photon's energy equals the kinetic energy the electron lost: \(h\nu = K_{\text{before}} - K_{\text{after}}\).
Why the spectrum is continuous
How much energy an electron loses in one braking depends on how close it passes the nucleus. A far pass takes a little, and a close pass takes a lot. Nothing picks out one special amount. So the photons from many electrons have every energy from zero up to the whole kinetic energy of an electron.
Most electrons do not give up their energy in one photon. An electron typically makes many collisions, losing a little energy each time. Most of these losses only shake the atoms of the target, and that is the 99 % that becomes heat. The few photons made along the way each take some part of the energy. Together they form a smooth band of wavelengths, the continuous spectrum.
The electrons that strike the target of a \(25\,\text{kV}\) tube all arrive with \(25\,\text{keV}\). Why do the X-rays have a spread of energies instead of one energy?
The shortest wavelength
A photon can take all of an electron's kinetic energy, but never more. So the largest photon energy from a tube at voltage \(V\) is \(eV\). Photon energy is \(hc/\lambda\), so the largest energy means the shortest wavelength. A tube gives no X-ray at all shorter than this cut-off wavelength.
\[ K = eV \]
Why. each electron crosses the tube voltage \(V\) from rest and arrives with kinetic energy \(eV\).
\[ h\nu_{max} = eV \]
Why. in the rare case where the electron is stopped in one braking, one photon takes all of \(K\).
\[ \dfrac{hc}{\lambda_{min}} = eV \]
Why. \(\nu = c/\lambda\), so the largest frequency is the shortest wavelength.
\[ \lambda_{min} = \dfrac{hc}{eV} \]
Read it. solve for the wavelength.
With \(hc = 1240\,\text{eV nm}\), the arithmetic is easy. An electron that crosses \(V\) volts carries \(V\) electron volts. So \(\lambda_{min} = 1240/V\) nanometres, with \(V\) in volts. At \(30\,\text{kV}\) this gives \(1240/30000 = 0.0413\,\text{nm}\).
Find the cut-off wavelength of an X-ray tube at (a) \(12.4\,\text{kV}\) and (b) \(31\,\text{kV}\). Give each in nm and in Å.
(a) \(\lambda_{min} = 1240/12400 = 0.100\,\text{nm} = 1.0\,\text{Å}\).
(b) \(\lambda_{min} = 1240/31000 = 0.040\,\text{nm} = 0.4\,\text{Å}\).
(c) The voltage grew by a factor of 2.5, and the cut-off shrank by the same factor: \(\lambda_{min} \propto 1/V\).
Values used: \(hc = 1240\,\text{eV nm} = 12400\,\text{eV Å}\), \(1\,\text{Å} = 0.1\,\text{nm}\).
An X-ray tube runs at \(40\,\text{kV}\). Find its cut-off wavelength in nm. Take \(hc = 1240\,\text{eV nm}\).
The cut-off depends on the tube voltage and on nothing else. Change the target from molybdenum to tungsten, and the cut-off stays where it is. Turn up the filament current, and it stays again. Only the energy of a single electron, \(eV\), can set the largest single photon.
A tube with a tungsten target gives a cut-off of \(0.041\,\text{nm}\). The target is replaced by molybdenum, and the voltage is unchanged. What is the new cut-off?
In 1915 William Duane and Franklin Hunt measured the shortest wavelength from an X-ray tube at several voltages. They found that \(\lambda_{min}\) times \(V\) stayed the same, exactly as \(\lambda_{min} = hc/eV\) says. From it they worked out Planck's constant. At \(30\,\text{kV}\) and \(0.0413\,\text{nm}\), \(h = eV\lambda_{min}/c\) gives \(6.6 \times 10^{-34}\,\text{J s}\). Their value agreed with the one from the photoelectric effect. The same \(h\) that frees an electron with light also limits the X-rays that an electron can make.
Reading the graph
An X-ray spectrum is drawn as a graph of intensity against wavelength. The graph has a clear shape. It is zero up to \(\lambda_{min}\), because no photon is shorter. It then rises steeply to a broad hump and falls away slowly at longer wavelengths. A few sharp, tall peaks may stand on top of the hump.
Move the slider in the figure above and watch three things. The cut-off slides to shorter wavelengths as the voltage rises. The hump moves left with it, because every photon can now be larger. The whole curve grows taller, because each electron brings more energy and a larger share becomes X-rays. Now switch the target. The cut-off does not move, but the tungsten curve is taller.
Each control of the tube leaves its own mark on the graph:
- Tube voltage up: \(\lambda_{min}\) and the hump move to shorter wavelengths, and the curve rises.
- Filament current up: the whole curve is multiplied by the same factor, and \(\lambda_{min}\) stays.
- Heavier target: the continuous curve rises, \(\lambda_{min}\) stays, and the sharp peaks move to new places.
A molybdenum tube is run at \(25\,\text{kV}\) and then at \(50\,\text{kV}\). The two spectra are drawn on one graph. One starts at \(0.050\,\text{nm}\) and the other at \(0.025\,\text{nm}\). Which curve belongs to which voltage? Check both cut-offs.
The cut-off is \(\lambda_{min} = 1240/V\) nm, which shrinks as \(V\) grows.
At \(25\,\text{kV}\): \(1240/25000\). At \(50\,\text{kV}\): \(1240/50000\).
\(25\,\text{kV}\) gives \(0.0496 \approx 0.050\,\text{nm}\). \(50\,\text{kV}\) gives \(0.0248 \approx 0.025\,\text{nm}\).
The curve starting at \(0.050\,\text{nm}\) is the \(25\,\text{kV}\) spectrum. The one starting at \(0.025\,\text{nm}\) is the \(50\,\text{kV}\) spectrum, and it is the taller.
Put these tubes in order of their cut-off wavelength, the longest first.
The voltage of an X-ray tube is doubled while the tube current is kept the same. Explain what happens to the cut-off wavelength and to the intensity.
Each electron now arrives with twice the energy, \(2eV\). The largest photon doubles in energy, so the cut-off wavelength \(hc/eV\) halves. The same number of electrons each second brings twice the energy, and a larger share of it becomes X-rays, so the intensity rises at every wavelength.
Why does the spectrum fall away at long wavelengths, although the braking makes many small photons? There are two reasons. First, a long-wavelength photon carries little energy, and a narrow range of photon energies is spread over a wide range of wavelengths. Second, soft X-rays are absorbed by the target itself and by the glass of the window before they escape.
Cancer hospitals, such as the Tata Memorial Hospital in Mumbai, treat deep tumours with X-rays from a linear accelerator. The machine speeds electrons up through about \(6\) million volts before they strike a tungsten target. The cut-off is \(1240/6000000 = 0.0002\,\text{nm}\). These very hard X-rays pass through the skin and deliver their energy deep inside the body.
- Thinking the cut-off wavelength depends on the target metal.\(\lambda_{min} = hc/eV\) depends on the tube voltage only. Change the target, and the cut-off stays.
- Writing \(\lambda_{min} = 1240/V\) with \(V\) in kilovolts and the answer in nm.With \(V\) in volts the answer is in nm. With \(V\) in kilovolts use \(1.24/V\) nm.
- Taking the peak of the hump as the cut-off.The cut-off is where the curve starts, at the shortest wavelength. The hump lies to its right.
- Thinking the filament current moves the cut-off.More current means more electrons, each with the same \(eV\). The curve grows, and the cut-off stays.