[theThinker (OP)]

What is meant by "gamma energies" in Compton scattering experiment using gamma rays. 

Energy calibration:
"A number of plastic disk sources of known gamma energies are supplied. Determining the
peak channel numbers for these known energies allows calibration of the equipment."

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

1) Does it mean the frequencies of the gamma rays could be obtained using: E=hf; f=E/h
2) If the frequency of the rays (e.g. Na-22) are measured directly with an instrument, would it give the same as f = E/h ?
3) What technique is there to directly measure frequency or wavelength?

[evan_au]

1) Yes. You could measure energy of gamma rays directly by a calorimeter, for example.
2) The frequency of gamma rays is very high. It is hard to measure this directly.
- With the newer optical clocks, metrologists do measure the frequency of visible light, but gamma rays are very much higher frequencies.
3) The wavelength of gamma rays is very small. It is hard to measure this directly.
- With X-Rays, it is possible to use the angle of diffraction from the atoms of a crystal to determine the wavelength, but gamma rays have very much shorter wavelengths.

I am sure alan calverd could add more.

[alancalverd]

Not a lot more, sadly!

Calorimetry will measure absorbed dose but not the energy of individual gammas. You can use crystal diffraction methods to determine gamma wavelength up to around 10 MeV, but at higher energies we are dependent on the point calibration of GeLi and similar spectrometers by calculating the energy loss from the mass loss of emitting nuclei.

The good news is that as far as we know, E = hf is true for all electromagnetic radiation (and why should it be otherwise?) so nobody bothers to measure frequency or wavelength directly above about 50 keV (i.e. the range of practical crystallography).

[Bored chemist]

Does it mean the frequencies of the gamma rays could be obtained using: E=hf; f=E/h

Yes
 the 511  KeV gammas have a frequency of 1.24 * 10^20 Hz

That's many millions of times higher than the fastest direct counters.

[theThinker (OP)]

Thanks.

My doubts now cleared.

[evan_au]

Calorimetry will measure absorbed dose but not the energy of individual gammas

This sparked a few random thoughts...

Could you estimate the energy of individual gamma rays by:
- Taking a gamma ray source of a steady (average) rate - like those calibration samples mentioned above
- Use a calorimeter to measure the total gamma ray energy coming out (as an average)
- Use a particle counter to measure the total rate of gamma rays coming out (as an average)
- Do a bit of arithmetic to determine the average energy of each gamma ray?

Unfortunately, if the source emits gamma rays of two different energies, this will just give you a weighted average of the gamma ray energy, not the actual spectrum.

What about using atomic recoil when an atom produces a gamma ray?
- If the source were embedded in plastic (like the calibration samples mentioned above), would this kick the atom enough so that it emitted Cerenkov radiation?
- A ring detector might be able to estimate the maximum velocity of the atom by the angle of the Cerenkov photons....
- Conservation of momentum would give the momentum of the photon
- But you still end up with some variant on E=hf (using Planck's constant) to determine the frequency (a quantum effect) from the momentum (a relativistic effect).

[alancalverd]

Could you estimate the energy of individual gamma rays by:- Taking a gamma ray source of a steady (average) rate - like those calibration samples mentioned above- Use a calorimeter to measure the total gamma ray energy coming out (as an average)- Use a particle counter to measure the total rate of gamma rays coming out (as an average)- Do a bit of arithmetic to determine the average energy of each gamma ray?

I've used a calorimeter to measure dose from a gamma source, but the flux required to produce a sensible temperature change (we measured 0.00001K/second to +/- 0.1%!) is too high to be counted as individual photons, and photon detector efficiency is itself energy-dependent. However I did use calorimetry to calibrate the energy of an electron accelerator where the second measurement (electron current) was fairly easy and completely independent of their energy.

[Bored chemist]

the flux required to produce a sensible temperature change (we measured 0.00001K/second to +/- 0.1%!) is too high to be counted as individual photons,

If you put the photon counter much further away and relied on the inverse square law...?

10µK/s is quite an impressive trick.

[happy6666]

You know my teacher (HS) said that gamma rays were mass-less yet I did the math and found that E = hf represents the units of the kinetic energy (kg m2/s2). My teacher say I am just a dumb cheerleader and I should do cheers and not challenge him. How should I answer this old toad?

[Kryptid]

You know my teacher (HS) said that gamma rays were mass-less yet I did the math and found that E = hf represents the units of the kinetic energy (kg m2/s2). My teacher say I am just a dumb cheerleader and I should do cheers and not challenge him. How should I answer this old toad?

Strictly speaking, photons do have relativistic mass due to E=mc².

[happy6666]

You know my teacher (HS) said that gamma rays were mass-less yet I did the math and found that E = hf represents the units of the kinetic energy (kg m2/s2). My teacher say I am just a dumb cheerleader and I should do cheers and not challenge him. How should I answer this old toad?

Strictly speaking, photons do have relativistic mass due to E=mc².

What does the variable E represent the energy of? I ask that question since mc² = hv doesn't quite make any sense to me but I may be wrong if I say so myself and I tried to ask the Toad but he has not answered my email but I asked my boyfriend and he said that I should just write what I got and wait for the toad to answer back.

[Kryptid]

What does the variable E represent the energy of?

Whatever the entity in question is (the photon in this case). Any mass from anywhere has an energy associated with it.

[evan_au]

E = hf represents the units of the kinetic energy (kg m2/s2)

Energy comes in many different forms, eg:
- Kinetic energy (moving objects)
- Potential energy (eg a cannonball at the top of the leaning tower of pisa)
- Chemical energy (eg a mixture of Hydrogen and Oxygen)
- Light Energy
- Nuclear energy
- Einstein even extended the concept of energy to mass-energy
- All of these forms of energy have the same units
- But it's not equally easy to turn one form of energy into another form of energy, so they are not directly equivalent

So, you can turn potential energy into kinetic energy by just dropping the cannonball
- But turning light into kinetic energy is not so easy
- So even though light energy and kinetic energy have the same units, it is not quite right to say that light has kinetic energy
- It is more correct to say that light has momentum

It just so happens that there is a satellite in orbit with a "solar sail"
- This turns the momentum of light into kinetic energy (but extremely inefficiently: Only a very small fraction of the incoming light energy is turned into kinetic energy!)
See: https://www.space.com/lightsail-2-solar-sail-mission-success.html

[alancalverd]

Beware of taking too naïve an interpretation of dimensions! Torque has the dimensions of energy, but not even Dr Who has a photon spanner (yet!)

E = mc^2 is best interpreted as the energy released in the annihilation of mass - google  "pair production" or "nuclear fission" for everyday uses of the phenomenon.

[yor_on]

It's like everything else Happy :)

It depends on how you define it. If you would ask Pete (who knows his relativity excellently) I'm pretty sure he might call it a equivalence to 'mass'.

But it's light, and light is in main stream physics defined to be 'mass less'.
But then you have the idea of 'photons in a perfectly reflecting box' giving the box a higher mass.
=

And no, whatever choice you make here it's yours. Don't let anyone tell you that you don't use your mind.
==

Maybe you should start to look into mathematics?
https://physics.stackexchange.com/questions/348892/does-a-photon-box-have-gravitational-mass/349056

[yor_on]

As it touches SR, as well as GR. Here's a book (SR- .pdf) free to read by Benjamin Crowell.  I've found him pretty straightforward on stack exchange. http://www.lightandmatter.com/sr/sr.pdf

[theThinker (OP)]

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 energy standards determined?

[evan_au]

How are these energy standards determined?

One aspect of the answer: They choose isotopes that are easy to produce, non-toxic, and produce gamma rays of a well-defined frequency at a reasonable rate for a reasonable period of time.

As for how the energy of these gamma rays was determined...
- These gamma rays are of sufficient energy that the difference in mass of the nuclei (decayed or not) could be distinguished by a high-resolution mass spectrometer.

[Bored chemist]

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 energy standards determined?

In the case of 22Na the photon is actually produced by annihilation of a positron, so the energy can be calculated from the mass of the electron which is accurately known.

[theThinker (OP)]

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 energy standards determined?

In the case of 22Na the photon is actually produced by annihilation of a positron, so the energy can be calculated from the mass of the electron which is accurately known.

Yes. I now notice the 511 KeV which is the energy of an electron.

I think evan_au may be wrong that the line energy is calculated based on the difference in mass in nuclear reactions. Gamma decay involves no mass changes; it comes from difference between the excited nuclear states going back to the ground states emitting a gamma ray photon.