High School What is the minimum voltage required to get characteristic spectrum

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SUMMARY

The minimum voltage required to produce a characteristic spectrum in X-ray generation using a Coolidge tube is determined to be approximately 15,500 volts (1.55 x 104 Volts). This calculation is based on the energy equation E=hc/λ, where λ is the wavelength of emitted radiation. The discussion highlights that the commonly referenced wavelength of 0.8 x 10-10 meters is incorrect; the accurate values for characteristic radiation wavelengths are 0.06323 nm and 0.07101 nm. The voltage must exceed the ionization energy of the K shell to effectively produce characteristic X-rays.

PREREQUISITES
  • Understanding of X-ray generation principles, specifically using a Coolidge tube.
  • Familiarity with the energy-wavelength relationship in quantum mechanics (E=hc/λ).
  • Knowledge of ionization energies, particularly for K shell electrons.
  • Basic grasp of bremsstrahlung and characteristic spectra in X-ray physics.
NEXT STEPS
  • Research the ionization energies of various elements, focusing on K shell electrons.
  • Study the differences between bremsstrahlung and characteristic X-ray production.
  • Explore advanced calculations involving X-ray wavelengths and energies.
  • Learn about the practical applications of Coolidge tubes in modern X-ray technology.
USEFUL FOR

Students and professionals in physics, particularly those specializing in X-ray technology, radiology, and materials science, will benefit from this discussion.

Samar A
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Hi,
The x-ray consists of the bremsstrahlung spectrum and the characteristic spectrum. We can get x-rays by using Coolidge tube where there are an applied voltage on the tube between the filament and the target. The bremsstrahlung is depeding on the PD between the filament and the target, but it is produced anyways. The characteristic spectrum requires a high PD to be produced.
I wonder how could we determine the less voltage is sufficient for the emergence of characteristic spectrum by using a diagram of the two spectra, like this:
KCyJ8.jpg

(Note: the information on the photo were not clear so I deleted them and wrote them again using my cellphone)
Anyways, in my textbook it tried to give the less voltage needed for the characteristic spectrum, and it did this:
E=hc/λ =(6.625*10^-34)*(3*10^8)/ (0.8*10^-10)
=2.48*10^-15 J
And then, because E=eV:
2.48*10^-15= (1.6*10^-19) *V
So, V= 1.55*10^4 Volts.

I am not convinced with the way it used and I don't know how to determine it myself, either.
Could someone explain me way it used the value (0.8*10^-10) in the first part of the solution, because I don't think it is the longest wavelength the characteristic would appear.
 
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Samar A said:
because I don't think it is the longest wavelength the characteristic would appear.
Why not? Characteristic radiation occurs as soon as the total energy of the electrons hitting the cathode is sufficient to excite the atoms in it. You can determine this energy based on the wavelength the atoms emit.
 
Isn't this "Elemtehan"? xD
I'm really happy to see another student in Egypt wants to study correctly
Ok... I know that it's an old topic and you no longer need it, but I think I can help. After googling it for a long time, I've found that you (and I) are right. It shouldn't be calculated this way. 0.08nm doesn't indicate anything to be used as characteristic spectrum wavelength. It's 0.06323nm and 0.07101nm. Also, even if it were the correct wavelength, the calculated energy would be the difference in energy between M and K and between L and K. The required voltage should be the ionization energy of K (as it is required to eject an existing electron and get another into its place to emit an X-ray photon)
I wish it helps.
 

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