Highest Frequency emitted by bremsstrahlung

In summary: I forgot to include the conversion into joules when I was doing the math. :)In summary, an electron is accelerated through a potential difference of 8.5 kV and emits electromagnetic waves at a frequency of 2.067 Hz.
  • #1
Blanchdog
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Homework Statement


Electrons are accelerated in a television tube through a potential difference of 8.5 kV. What is the highest frequency of the electromagnetic waves emitted when these electrons strike the screen of the tube?

Homework Equations


λ = c/f
E (of the electron) = V*e
E (of the photon) = h*c/λ

The Attempt at a Solution



Without knowing the work function of the tube, the best I could come up with was to assume that all the kinetic energy of the electron was transferred to the photon and set the energies equal to one another. Simple algebra substituting the wavelength for it's definition in terms of frequency gives the equation:

f = V*e/h​

Substitution of numeric values then gives

f = 8500*1.6E-19/6.58E-16 = 2.067 Hz

EDIT: I accidentally used h-bar instead of h. It still comes out different from the desired answer though.
f = 8500*1.6E-19/4.14E-15 = .329 Hz​

The online homework I have says that I am off by 18 orders of magnitude: the correct answer is 2.1E18

Where am I going wrong here?
 
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  • #2
Blanchdog said:
f = V*e/h​
Substitution of numeric values then gives

f = 8500*1.6E-19/6.58E-16 = 2.067 Hz​
I think you forgot the conversion from eV to Joules for the electron's initial energy...?
 
  • #3
Units. Always write out the units.
 
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  • #4
berkeman said:
I think you forgot the conversion from eV to Joules for the electron's initial energy...?
He did not. He inserted the Planck constant in units of eV s instead of in SI units when everything else is in SI units.
 
  • #5
Orodruin said:
He did not. He inserted the Planck constant in units of eV s instead of in SI units when everything else is in SI units.
Ah, thanks!
 
  • #6
berkeman said:
I think you forgot the conversion from eV to Joules for the electron's initial energy...?

The dimensional analysis works out, currently the units are eV/(eV*s), leaving 1/s = Hz. I did however accidentally use h-bar instead of h, so fixing that:

f = 8500*1.6E-19/4.14E-15 = .329 Hz​
 
  • #7
Blanchdog said:
The dimensional analysis works out, currently the units are eV/(eV*s)
Dimensional analysis is about getting the right physical dimension. Your problem is with units, not with dimensions. If you want to use eV as your energy unit, you will be using units of the elementary charge for charges. Your numerator with the numerical values you inserted have the units V C (i.e., Joules), not eV.
 
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  • #8
Orodruin said:
Dimensional analysis is about getting the right dimension Your problem is with units, not with dimensions. If you want to use eV as your energy unit, you will be using units of the elementary charge for charges. Your numerator with the numerical values you inserted have the units V C (i.e., Joules), not eV.

Ahhhhh! Thank you, you're exactly right.
 
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What is the highest frequency emitted by bremsstrahlung?

The highest frequency emitted by bremsstrahlung is determined by the energy of the incident electrons and the atomic number of the target material.

What is the relationship between the incident electron energy and the highest frequency emitted by bremsstrahlung?

The higher the energy of the incident electrons, the higher the maximum frequency of the emitted bremsstrahlung radiation.

Can the highest frequency emitted by bremsstrahlung be controlled?

Yes, the maximum frequency can be controlled by adjusting the energy of the incident electrons and the atomic number of the target material.

What materials are commonly used as targets for bremsstrahlung radiation?

Heavy elements such as tungsten, gold, and lead are commonly used as targets for bremsstrahlung radiation due to their high atomic numbers and ability to produce high-energy photons.

What is the significance of the highest frequency emitted by bremsstrahlung?

The highest frequency emitted by bremsstrahlung is important in medical imaging, material analysis, and other applications where high-energy photons are needed to penetrate through dense materials.

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