Determine X-ray Wavelength & Frequency

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SUMMARY

The discussion focuses on calculating the wavelength and frequency of x-rays emitted when 100-keV electrons strike a target. The calculations yield a wavelength of 0.12 Å and a frequency of 2.4 x 1019 Hz, correcting an initial typo in the frequency exponent. The participants emphasize the importance of using electron volts (eV) for energy calculations and remembering the constant hc = 1240 eV-nm for conversions between wavelength and photon energy. The problem is framed within the context of the photoelectric effect, noting that the work function is negligible compared to the energy of the electrons.

PREREQUISITES
  • Understanding of kinetic energy and photon energy relationships
  • Familiarity with the photoelectric effect and its implications
  • Knowledge of energy units, specifically electron volts (eV) and joules (J)
  • Basic grasp of wavelength and frequency calculations in physics
NEXT STEPS
  • Learn how to convert between energy units, specifically eV and J
  • Study the photoelectric effect and its applications in modern physics
  • Explore the relationship between wavelength and frequency using the equation c = λf
  • Investigate the significance of the constant hc = 1240 eV-nm in photon energy calculations
USEFUL FOR

Physics students, researchers in quantum mechanics, and professionals working with x-ray technology or photon energy calculations will benefit from this discussion.

Von Neumann
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Problem:

Determine the wavelength and frequency of the emitted x-rays when 100-keV electrons strike a target.

Solution:

Assuming all kinetic energy of electrons is used to produce the x-rays,

E_initial=E_final
K+m_0*c^2=hf+m_0*c^2
K=hf
K=hc/λ
=>λ=hc/K=(6.63x10^-34 Js)(3.00x10^8 m/s)/(1.60x10^-14 J) [because 100 keV =1.60x10^-14 J]

λ=0.12 Å

Also since K=hf,

f=K/h=(1.60x10^-14 J)/(6.63x10^-34 Js)=2.4x10^-19 Hz

Is this correct?
 
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Looks essentially correct. Just recheck the exponent on your frequency (and my apologies if that's just a simple typo on your part)

I am wondering if this might have been intended as a "reverse photoelectric effect" problem. But since the work function is negligible compared to 100 keV, you come up with the same K≈hf to work with.

A few tips for working problems like this:

It goes easier if you can get used to working in eV rather than always converting to joules.

A useful constant to remember, for going between wavelength and photon energy, is hc=1240 \text{ eV-nm}

It's a good habit to think about whether your answer to any problem makes sense. For example, we (hopefully) know that visible light has a frequency way higher than 1 Hz, and x-rays have frequencies way higher than visible light. Again, my apologies if the 10^-19 was simply a typo.
 
Oh yes, that is a typo! Should be x10^19~My apologies, but I'm happy you caught it.

Thanks for the tips, and I'll certainly put in time to commit the value for hc to memory.
 

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