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 Nuclear and Particle Physics Nuclear and Particle Physics Help Forum Jul 26th 2010, 11:23 PM #1 Senior Member   Join Date: Feb 2009 Location: Singapore Posts: 206 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! __________________ To each his own.   Sep 2nd 2010, 07:19 PM   #2
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 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?   Sep 2nd 2010, 09:14 PM   #3
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 Originally Posted by sa-ri-ga-ma 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
__________________
To each his own.   Sep 2nd 2010, 11:04 PM   #4
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 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)   Sep 3rd 2010, 03:52 AM #5 Senior Member   Join Date: Feb 2009 Location: Singapore Posts: 206 Can you tell me what the symbols are? the Y, p. Thanks __________________ To each his own.   Sep 3rd 2010, 06:44 AM   #6
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 Can you tell me what the symbols are? the Y, p. Thanks
p is momentum and γ = 1/ sqrt(1- v^2/c^2)   Sep 19th 2010, 09:27 PM #7 Senior Member   Join Date: Feb 2009 Location: Singapore Posts: 206 I still can't get it, the what is the Y, as in volts? or something like that? I'm confused thanks __________________ To each his own.  Tags electron, refraction Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post Germanlongstrike Light and Optics 13 Apr 24th 2017 01:06 AM DarkSoul Light and Optics 2 Dec 26th 2010 06:22 PM {smile} Waves and Sound 2 Aug 19th 2009 08:44 PM iEricKim Light and Optics 2 May 19th 2009 01:41 AM Air Light and Optics 3 May 13th 2008 05:12 AM