Bremsstrahlung wavelength range

In summary, the conversation discusses the efficiency of X-ray tubes in turning electrical energy into X-ray energy, and the potential use of synchrotron radiation as a more efficient alternative. The conversation also touches on the idea of using hard drive disks and magnets to produce MeV radiation, but it is determined that this is not a feasible option. The conversation concludes with a discussion on the desired energy level for a specific experiment and the potential use of a synchrotron with 400 MeV electrons and a 15-meter diameter to output 4 MeV photons.
  • #1
pixelpuffin
45
1
i need to calculate what percentage of the energy put into a X-ray tube will be between the minimum wavelength and the longest wavelength suitable for it's application (as to calculate how efficiently it does it's job) with only the input voltage (unless the anode material is important)
the input voltage of the X-ray tube is >4,000,000 volts (because the minimum energy photon i need is 4MeV)
the wavelength range is 4MeV or 3.1e-7 micrometers or 9.6e11 Ghz to the input voltage
the efficiency required is 90%

also at what voltage will i start to have electron soar right into the nucleus and turn a proton into a neutron because that would cause problems (though it seems in most cases they synthesised nucleus would quickly decay back spitting out an electron in the process)
 
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  • #2
X-ray tubes are typically less than 1% efficient at turning electrical energy into X-ray energy. Most of the input energy ends up as heat. I think an efficiency of 90% is out of the question. Perhaps you should look into synchrotron radiation, which can be much more efficient.
 
  • #3
i meant the efficiency after losses to heat but ill also look into that
 
  • #4
after some research on free electron lasers and synchrotrons i found that with magnets i could make i'd need 19.7 GeV electrons
i also found two potential options though
if i could use hard drive disks with back and forth 1s to 0s i would only need 62 MeV electrons (assuming the magnets are 10 nanometers across)
alternatively i could have an incredibly powerful magnet at the end of a vacuum tube that turns the electrons very suddenly however I'm not sure how to calculate the electron energy required
i would guess that it would act like a single very small magnet (how far the electron gets by the midpoint of the turn?) and thus could be calculated with either the same or a similar equation
 
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  • #5
phyzguy said:
X-ray tubes are typically less than 1% efficient at turning electrical energy into X-ray energy. Most of the input energy ends up as heat. I think an efficiency of 90% is out of the question. Perhaps you should look into synchrotron radiation, which can be much more efficient.
Synchrotron radiation for 4 MeV? I didn't see that so far.

if i could use hard drive disks with back and forth 1s to 0s i would only need 62 MeV electrons (assuming the magnets are 10 nanometers across)
They are not, and they are not strong enough for significant radiation. It would also mean your beam would need a focus of a few nanometers (to see those magnets at all).
If you want to get MeV radiation out of electrons, you'll need the intense electric fields around atoms, which means you are back to the x-ray tube principle.

Why do you need 4 MeV?
You can get up to 2.5 MeV with radioactive sources.
 
  • #6
mfb said:
If you want to get MeV radiation out of electrons, you'll need the intense electric fields around atoms, which means you are back to the x-ray tube principle.

Why do you need 4 MeV?
You can get up to 2.5 MeV with radioactive sources.

actually after some research i found i could make a synchrotron output 4 MeV with a mere 15 meter diameter and 400 MeV electrons (synchrotron in question can handle up to 1 GeV)

2.5 Mev isn't high enough energy for the experiment i want to do
 
  • #7
pixelpuffin said:
actually after some research i found i could make a synchrotron output 4 MeV with a mere 15 meter diameter and 400 MeV electrons (synchrotron in question can handle up to 1 GeV)

2.5 Mev isn't high enough energy for the experiment i want to do

Are you sure? After researching it a little more I think that mfb is right that 4 MeV is too high a photon energy for most synchrotron sources. How did you decide that 400 MeV electrons and a 15 m radius will output 4 MeV photons?
 
  • #8
pixelpuffin said:
actually after some research i found i could make a synchrotron output 4 MeV with a mere 15 meter diameter and 400 MeV electrons (synchrotron in question can handle up to 1 GeV)
How, where?

2.5 Mev isn't high enough energy for the experiment i want to do
What do you want to do?
 

Related to Bremsstrahlung wavelength range

1. What is Bremsstrahlung radiation?

Bremsstrahlung, also known as braking or deceleration radiation, is a type of electromagnetic radiation emitted by charged particles, such as electrons, when they are accelerated by a strong electric field or decelerated by the Coulomb force of an atomic nucleus.

2. What is the wavelength range of Bremsstrahlung radiation?

The wavelength range of Bremsstrahlung radiation can vary widely, depending on the energy of the charged particle and the strength of the electric field. Generally, it can range from a few nanometers to several hundred nanometers.

3. How is Bremsstrahlung radiation produced?

Bremsstrahlung radiation is produced when charged particles, such as electrons, are accelerated or decelerated by an electric field. This can occur in a variety of situations, such as in particle accelerators, X-ray tubes, and even lightning.

4. What are the applications of Bremsstrahlung radiation?

Bremsstrahlung radiation has a wide range of applications, including medical imaging (X-rays), industrial imaging and inspection, and materials analysis. It is also used in particle physics research to study the properties of charged particles and their interactions with matter.

5. How does the energy of Bremsstrahlung radiation relate to its wavelength?

The energy of Bremsstrahlung radiation is inversely proportional to its wavelength. This means that as the wavelength decreases, the energy of the radiation increases. This relationship is described by the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength.

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