Calculating Electron Diffraction in a 50.0 kV Television

[marc.orr]

question:
Electrons in a television are accelerated through a potential difference of 50.0 kV and then pass through a deflecting capacitor of width 1.00 cm. Are electron diffraction effects important in this situation? Justify your answer with one or more calculations.

A couple of questions about the homework problem above that I was assigned.

1) At the link below, they show the formula for the relativistic speed of an electron accelerated through a potential differnce. I can't seem to derive it. I just set KE(e) = PE(field)
E-Eo = eV ; E = relativistic energy, Eo = rest energy

http://en.wikipedia.org/wiki/Electron_diffraction

2) In the problem he gives the width of the capacitor. I'm guessing this is supposed to be some sort of diffraction grading. Do you need to use this information. I was just thinking that you could compare the debroglie wavelength the to diameter of the electron.

-Marc

 

[Jhero]

Hello Marc,

Thank you for your questions regarding the homework problem. I can help clarify the concepts and provide some calculations to justify the importance of electron diffraction effects in this situation.

Firstly, let's address your question about deriving the formula for the relativistic speed of an electron accelerated through a potential difference. This formula can be derived using the conservation of energy principle, where the kinetic energy gained by the electron is equal to the potential energy it gains from the electric field. We can express this as:

KE = PE
(1/2)mv^2 = eV

Where m is the mass of the electron, v is its velocity, e is the elementary charge, and V is the potential difference. From here, we can rearrange the equation to solve for v:

v = √(2eV/m)

This is the formula for the relativistic speed of the electron.

Moving on to your second question about the width of the deflecting capacitor, this information is indeed important in considering electron diffraction effects. In order for diffraction to occur, the width of the barrier or grating must be comparable to the wavelength of the particles. In this case, the width of the capacitor is 1.00 cm, while the electron's wavelength can be calculated using the de Broglie equation:

λ = h/mv

Where h is Planck's constant, m is the mass of the electron, and v is its velocity. Substituting the value for v from our previous calculation, we get:

λ = h/√(2eV/m)

Using the values for h, e, and V, we can calculate the wavelength of the electron. If this value is comparable to the width of the capacitor, then we can say that electron diffraction effects are important in this situation.

I hope this helps clarify the concepts and provides some calculations to justify the importance of electron diffraction effects in this scenario. Let me know if you have any further questions. Good luck with your homework!