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Old Jul 26th 2010, 11:23 PM   #1
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Electron refraction

A parallel beam of electrons is moving with velocity v is incident normally on a thin graphite film of atomic spacing 1.2*10^-10 m. The beam is diffracted through an angle of 11 degrees, where 2dsin theta=lambda, the wavelength value.
Calculate the wavelength, velocity v and the accelerating voltage needed to produce this velocity.

I found the answers to the first two correctly, 4.6*10^-11m and 1.4*10^7m/s respectively.
I used eV=1/2 m v^2, to find the voltage.
V=(1/2*10^-31*(1.4*10^7)^2)/1.6*10^-19=612.5V answer is supposed to be 450V
Thanks!
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Old Sep 2nd 2010, 07:19 PM   #2
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Originally Posted by arze View Post
A parallel beam of electrons is moving with velocity v is incident normally on a thin graphite film of atomic spacing 1.2*10^-10 m. The beam is diffracted through an angle of 11 degrees, where 2dsin theta=lambda, the wavelength value.
Calculate the wavelength, velocity v and the accelerating voltage needed to produce this velocity.

I found the answers to the first two correctly, 4.6*10^-11m and 1.4*10^7m/s respectively.
I used eV=1/2 m v^2, to find the voltage.
V=(1/2*10^-31*(1.4*10^7)^2)/1.6*10^-19=612.5V answer is supposed to be 450V
Thanks!
How did you calculate the velocity? Since the velocity is close to the velocity of light, have you tried the relativistic approach?
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Old Sep 2nd 2010, 09:14 PM   #3
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Originally Posted by sa-ri-ga-ma View Post
How did you calculate the velocity? Since the velocity is close to the velocity of light, have you tried the relativistic approach?
You mean use E=mc^2? then E=eV?

m=9.11*10^(-31)
E=9.11*10^(-31)*[1.4*10^7]^2=1.8*10^(-16)
V=E/e=1.8*10^(-16)/1.6*10^(-19)=1116V
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Old Sep 2nd 2010, 11:04 PM   #4
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Originally Posted by arze View Post
You mean use E=mc^2? then E=eV?

m=9.11*10^(-31)
E=9.11*10^(-31)*[1.4*10^7]^2=1.8*10^(-16)
V=E/e=1.8*10^(-16)/1.6*10^(-19)=1116V
No.

The formula of diffraction is

dsinθ = mλ.

Ιf the velocity is comparable with the velocity of the light, the de Broglie relation becomes

γp = h/λ, where γ = 1/sqrt(1 - v^2/c^2)
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Old Sep 3rd 2010, 03:52 AM   #5
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Can you tell me what the symbols are? the Y, p.
Thanks
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Old Sep 3rd 2010, 06:44 AM   #6
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Originally Posted by arze View Post
Can you tell me what the symbols are? the Y, p.
Thanks
p is momentum and γ = 1/ sqrt(1- v^2/c^2)
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Old Sep 19th 2010, 09:27 PM   #7
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I still can't get it, the what is the Y, as in volts? or something like that? I'm confused
thanks
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