In operating scintillation gamma ray detectors, certain gamma ray standards are used for calibration:

Energy (KeV)
Na-22, 511
Mn-54, 835
Co-57, 122

How are these energies determined.

  • $\begingroup$ Are you asking how we know the Na22 line is 511keV, or asking what mechanism caused that energy to be emitted? $\endgroup$ – Bob Jacobsen Aug 4 '19 at 21:04
  • $\begingroup$ They may have used some sort of single-crystal x-ray spectrograph to measure the x-ray photon energies. See the article en.wikipedia.org/wiki/X-ray_spectroscopy and the section on "Early history of X-ray spectroscopy in the U.S." in it. $\endgroup$ – user93237 Aug 4 '19 at 23:45

Besides the szintillation gamma-ray detectors there are the semiconductor-based gamma-ray detectors. These semiconductor-based detectors (and especially the germanium-based one) provide a much better energy resolution in comparison to the szintillation detectors.

From Gamma-ray spectrometer: How a GRS works

Another approach relies on using Germanium detectors - a crystal of hyperpure germanium that produces pulses proportional to the captured photon energy; while more sensitive, it has to be cooled to a low temperature, requiring a bulky cryogenic apparatus.

Therefore you can calibrate a szintillation detector with the gamma energies measured with a semiconductor-based detector.

But this immediately raises the next question:
How can a semiconductor-based gamma-ray detector be calibrated?

For low-energy gamma-rays you can measure their wavelength $\lambda$ from diffraction by crystals. This is the same method as used for measuring X-ray wavelengths (see also Bragg's law), because low-energy gamma-rays and high-energy X-rays are essentially the same thing. From the wavelength $\lambda$ the energy $E$ of the gamma-photon can be calculated by $E = h\nu = hc/\lambda$.
However, this method is applicable only if the wavelength is not much smaller than the atomic distances in the crystal (i.e. for low-energy gamma-rays), otherwise the diffraction angles would be unmeasurable small.

  • $\begingroup$ But you still haven’t said how the line energies are established, and germanium detectors have nasty nonlinear energy response especially in the low energies the OP is asking about. $\endgroup$ – dmckee --- ex-moderator kitten Aug 4 '19 at 22:26
  • $\begingroup$ @dmckee Good point! I've extended my answer. $\endgroup$ – Thomas Fritsch Aug 5 '19 at 7:13
  • $\begingroup$ But how is "pulses proportional to the captured photon energy" experimentally verified as a valid relation between voltage output and energy input. $\endgroup$ – itsme Aug 6 '19 at 14:32
  • $\begingroup$ @itsme This can be verified by checking that the pulses (measured with the gamma photon detector) are proportional to $1/\lambda$ (with $\lambda$ measured with the diffraction method). See the formula $E=hc/\lambda$ in my answer. $\endgroup$ – Thomas Fritsch Aug 6 '19 at 14:47
  • $\begingroup$ But diffraction method not applicable for some higher energy gamma rays. Then the calibration works only for some lower energy levels. Can we assume extrapolation will also be strictly linear. $\endgroup$ – itsme Aug 6 '19 at 15:31

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