A photon with wavelength .0268 nm strikes an electron and is scattered at an angle of 30 degrees.
What is the de Broglie wavelength of the electron after the collision.
I found the scattered wavelength of the photon to be 2.713x10^-11m and the energy is 7.33X10^-15 J.
I can't figure out how to find the de Broglie wavelength of the electron, however. Thanks in advance.
What is the de Broglie wavelength of the electron after the collision.
I found the scattered wavelength of the photon to be 2.713x10^-11m and the energy is 7.33X10^-15 J.
I can't figure out how to find the de Broglie wavelength of the electron, however. Thanks in advance.
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Conservation of momentum: momentum p of the photon before the scattering event must equal the combined momentum of the electron and photon after the event.
Before:
p=h/λ₁
After:
p=h/λ₂ + pe
Where the 1 subscript denotes initial wavelength of the photon, 2 is final, and pe is the momentum of the electron. And so, by conservation of momentum, h/λ₁ = h/λ₂ + pe. Solving for the pe:
pe = h/λ₁ - h/λ₂
And we know the de Broglie wavelength of the electron is then
λ of electron = h/(h/λ₁ - h/λ₂) = λ₁λ₂/(λ₂ - λ₁)
Before:
p=h/λ₁
After:
p=h/λ₂ + pe
Where the 1 subscript denotes initial wavelength of the photon, 2 is final, and pe is the momentum of the electron. And so, by conservation of momentum, h/λ₁ = h/λ₂ + pe. Solving for the pe:
pe = h/λ₁ - h/λ₂
And we know the de Broglie wavelength of the electron is then
λ of electron = h/(h/λ₁ - h/λ₂) = λ₁λ₂/(λ₂ - λ₁)