Highest Frequency emitted by bremsstrahlung

Click For Summary
SUMMARY

The discussion centers on calculating the highest frequency of electromagnetic waves emitted by electrons accelerated through a potential difference of 8.5 kV in a television tube. The correct formula to use is f = V*e/h, where V is the potential difference, e is the elementary charge, and h is Planck's constant. The initial miscalculation resulted from using the wrong units for Planck's constant, leading to an incorrect frequency of 0.329 Hz instead of the correct value of 2.1E18 Hz. Proper unit conversion and dimensional analysis are crucial for accurate results in such calculations.

PREREQUISITES
  • Understanding of electromagnetic wave properties
  • Familiarity with the concepts of potential difference and energy conversion
  • Knowledge of Planck's constant and its units
  • Basic skills in dimensional analysis
NEXT STEPS
  • Research the conversion between electronvolts (eV) and joules for energy calculations
  • Learn about the implications of using different units in physics equations
  • Study the principles of bremsstrahlung radiation in detail
  • Explore advanced topics in quantum mechanics related to photon emission
USEFUL FOR

Students in physics, educators teaching electromagnetic theory, and professionals working in fields involving particle acceleration and radiation emission.

Blanchdog
Messages
56
Reaction score
22

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?
 
Last edited:
Physics news on Phys.org
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...?
 
Units. Always write out the units.
 
  • Like
Likes   Reactions: berkeman
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.
 
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!
 
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​
 
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.
 
  • Like
Likes   Reactions: Blanchdog
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.
 
  • Like
Likes   Reactions: berkeman

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
1K
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 34 ·
2
Replies
34
Views
3K
Replies
4
Views
2K