TV Cathode Ray Tube Consider a potential

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SUMMARY

The discussion centers on the calculations related to the operation of a cathode ray tube (CRT) television, specifically focusing on the kinetic energy of electrons accelerated by a potential of 15 kV. The kinetic energy is calculated to be 2.4 x 10-15 J, leading to a wavelength of 9.97 x 10-12 m for the electrons. Additionally, the wavelength of the emitted photons, calculated to be 8.28 x 10-11 m, falls outside the visible spectrum, indicating that CRT screens can emit x-rays due to the energy exchange process. This highlights the complexity of phosphorescence in CRT technology.

PREREQUISITES
  • Understanding of kinetic energy calculations in physics
  • Familiarity with the principles of wave-particle duality
  • Knowledge of photon energy equations (E = hf)
  • Basic concepts of cathode ray tube (CRT) operation
NEXT STEPS
  • Research the principles of wave-particle duality in quantum mechanics
  • Learn about the emission of x-rays from cathode ray tubes
  • Explore the phenomenon of phosphorescence and its applications
  • Study the effects of electron acceleration in vacuum tubes
USEFUL FOR

Students and educators in physics, electrical engineers, and professionals involved in the design and operation of cathode ray tube technology.

Tufts
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Homework Statement


In an old TV cathode ray tube, electrons are accelerated by a potential of 15 kV between the gun and screen. Find:

a) Kinetic energy of the electrons when they reach the screen
b) Its respective wavelength
c) Wavelength of the photons with maximum energy that are emitted from the screen.

Homework Equations



E = hf (Energy of a photon with frequency f)
f = c/\lambda (\lambda is the wavelength of respective photons)

\lambda = h/p (wavelength of electron depending on p, its momentum)

The Attempt at a Solution


a) E = e*V = 15000 eV = 15000 * 1.6 x 10-19 = 2.4 x 10-15 J

b) \lambda = h/p
But p = mv (where m = 9.1 x 10-31 kg)
We also know that 1/2 * mv2 = E
So, v = (2/m * E)1/2 => v = (2/(9.1 x 10-31)*2.4 x 10-15)1/2 = 0.73 x 108 m/s (not close enough to c, so we can use our electromagnetism approximation)

Therefore, \lambda = h/p = 6.626 * 10-34 / (9.1 x 10-31 * 0.73 x 108) = 9.97 x 10-12 m

c) Let us suppose that there is 100% efficient exchange of energy between electron and photon. E = hf => f = E/h => f = (15000 * 1.6 x 10-19)/(6.626 x 10-34) = 3.62 x 1018 Hz

But \lambda = c/f => \lambda = 8.28 x 10-11 m



To me, I am using the formulas correctly... but the wavelength of light on part C is not in the visual spectrem. Since we are talking about a TV here... that certainly does not make sense! What am I doing wrong?? Thanks before hand for all the help.

Tufts..
 
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Tufts said:
But \lambda = c/f => \lambda = 8.28 x 10-11 m



To me, I am using the formulas correctly... but the wavelength of light on part C is not in the visual spectrem. Since we are talking about a TV here... that certainly does not make sense! What am I doing wrong??
Nothing. CRT screens can and do emit a small amount of xrays. Phosphorescence is a complicated phenomenon by which energy of the electrons stimulates the target molecules to produce visible light.

AM
 
Thanks a lot, makes sense... I do remember the professor talking about x-rays emitted from CRT screens.

Tufts..
 

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