Unique λ at which X-ray and electron have same energy

In summary, the conversation discusses deriving an equation to show that there is a unique wavelength at which X-ray photons and electrons have the same energy. The relevant equations for photons and electrons are provided, and the attempt at a solution is presented. It is pointed out that the equation does not have a unique solution, but it is clarified that the goal is to show that the wavelengths intersect. Finally, the correct equation for the intersect is given as λ = 2cm/h.
  • #1
catkin
218
0

Homework Statement


Derive an equation to show that there is a unique wavelength at which X-ray photons and electrons have the same energy. Calculate this wavelength and energy.

Homework Equations


Here's what I thought was relevant. There may be others!

For photons: E = hf = h c / λ

For electrons:
λ = h / p (de Broglie)
K.E. = ½ mv^2
p = mv

The Attempt at a Solution


p = √(2mE)
Substituting in de Broglie
λ = h / √(2mE)
E = h^2 / 2mλ^2

Equating energies
(h c / λ)X-ray = (h^2 / 2mλ^2)electron
Gathering the constants (c, h, 2 and m -- the rest mass of an electron)
(λ)X-ray = (h / 2cm) * (λ^2)electron

... which does not have a unique solution :-(
 
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  • #2
Doesn't (λ)X-ray = (λ)electron?
 
  • #3
Thanks Borek :)

(nice hair!)

That's what the question asks the answerer to show so I hope it's true! I don't think I've shown that it is so.

Writing y for (λ)X-ray, x for (λ)electron and lumping the constants together as k my attempt shows, when the X-ray and electron energies are the same,
y = k x^2

Mmm ... the more I read the question the more ambiguous it becomes. Perhaps it would help if you could translate the question into unambiguous language.

Best

Charles
 
  • #4
Why do you still use two variables for wavelength, when you should use one?
 
  • #5
Thanks again, Borek.

I need to show that λ = fe(E) and λ = fp(E) intersect before that is legitimate (subscript e for electron, p for photon).

With that requirement now clear ...

For photons:
λ = hc / E
= (hc) * (1 / E)

For electrons:
λ = sqrt(h^2 / 2mE)
= sqrt(h^2 / 2m) * sqrt(1 / E)

Regardless of the constant values, these functions intersect once.

At the intersect the previous equation becomes (thanks!)
λ = (h / 2cm) * (λ^2)
= 2cm / h

and the rest is trivial.
 

What is the significance of the unique λ at which X-ray and electron have same energy?

The unique λ, or wavelength, at which X-ray and electron have the same energy is known as the Bragg wavelength. This phenomenon is significant because it allows for the use of X-rays to determine the atomic and molecular structure of materials.

How is the Bragg wavelength calculated?

The Bragg wavelength can be calculated using the Bragg equation, which is λ = 2d sinθ, where d is the distance between atomic planes in a crystal lattice and θ is the angle of incidence of the X-ray or electron beam.

What factors affect the unique λ at which X-ray and electron have same energy?

The unique λ at which X-ray and electron have the same energy is affected by the crystal structure, lattice spacing, and angle of incidence of the X-ray or electron beam. It can also be affected by the energy of the X-ray or electron beam.

How is the Bragg wavelength used in X-ray crystallography?

In X-ray crystallography, the Bragg wavelength is used to determine the position and spacing of atoms in a crystal lattice. By measuring the angle and intensity of X-rays diffracted from a crystal, scientists can calculate the Bragg wavelength and use it to determine the atomic structure of the material.

What other applications use the unique λ at which X-ray and electron have same energy?

The unique λ at which X-ray and electron have the same energy is also used in electron diffraction techniques, such as electron microscopy and electron diffraction spectroscopy. It is also used in X-ray fluorescence spectroscopy, X-ray diffraction imaging, and other analytical techniques to study the composition and structure of materials.

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