Electrons and Matter Waves

  • Thread starter nnof55
  • Start date
  • #1
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Homework Statement


A stream of protons, each with a speed of 0.8250c, are directed into a two-slit experiment where the slit separation is 2.00 10-9 m. A two-slit interference pattern is built up on the viewing screen. What is the angle between the center of the pattern and the second minimum (to either side of the center)?

Homework Equations


p = mv
lambda = h/p
d*sin(theta) = (m + 1/2)*lambda

The Attempt at a Solution


I have tried to first solve for the wavelength in the experiment by using p = mv. With this I get:
p = (1.673E-27)*(0.8250)*(3E8)
p = 4.14E-19

Then I solve for the wavelength using lambda = h/p:
lambda = (6.63E-34) / (4.14E-19)
lambda = 1.6E-15

Once I have the wavelength, I use the double slit formula from Young's Experiment to try and calculate the angle, by using m = 1 and then solving for arcsin:
theta = arcsin ( m*lambda / d)
theta = arcsin ( 1.5*(1.6E-15) / (2E-9))

However this gives me a very small angle which obviously is the incorrect answer.

Am I approaching this completely wrong, or am I just goofing up somewhere?
 

Answers and Replies

  • #2
166
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Welcome to PF!

For protons at 0.825c, you might want to calculate the relativistic momentum
 
  • #3
2
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Thanks! I was able to calculate the correct angle using the relativistic momentum equation!
 
  • #4
166
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Also, a small angle is not necessarily incorrect. To obtain large angles, the slit separation has to be comparable to the wavelength of the wave. Particles with large momenta will have very small wavelengths.
 

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