Wave-Particle Duality

  • #1

Homework Statement


At what Kinetic energy will a particle's debroglie wavelength equal its Compton Wavelength


Homework Equations


DeBroglie
λ = h/mv

Compton
λ = h/mc

The Attempt at a Solution



Setting the two equations equal to eachother, I got v = c, then said KE = (1/2)mc^2, but somehow that just doesn't sound right. What do you think?
 

Answers and Replies

  • #2
vela
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Nope, that's not right, as you suspected. The DeBroglie wavelength is actually [itex]\lambda = h/p[/itex] where p is the momentum of the particle. You need to use the relativistic expression for the kinetic energy to get the correct answer.
 
  • #3
So, it's when p = mc, so would you use E = Sqrt((pc)^2+(mc^2)^2) = sqrt((mc)^2+(mc^2))?
 
  • #4
vela
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Essentially, yes, but you need to get the algebra right. You have [itex]p = mc[/itex] so [itex]pc = mc^2[/itex] and
[tex]E=\sqrt{(pc)^2+(mc^2)^2} = \sqrt{(mc^2)^2+(mc^2)^2} = \cdots[/tex]Also, remember E gives the total energy, not the kinetic energy.
 

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