(adsbygoogle = window.adsbygoogle || []).push({}); Incident Photon λ = 0.1050*10^-9

Angle which photon is scattered (θ) = 60° (relative to intiial direction)

What angle is the electron scattered relaive to the original direction? (x)

Okay, so using linear momentum conservation:

(1) Incident photon momentum (p) = electron momentum (pe) * cos (x) + scattered photon momentum (q) * cos (60)

And (where λ' is the wavelength after scattering):

λ-λ' = h/mc (1-cosθ),

so pe*cos (x) = h/λ - h/λ'*2

pe*cos(x) = 3.2966*10^-24 (5sf)

And then usuing the scalar product of : pe = p - q to find pe:

eq. (2) :pe^2 = p^2+q^2 - 2pqcosθ

pe = 6.28*10^-24 (3sf)

Subbing this back into eq (1):

6.28*10^-24 (cos(x)) = 3.2966*10^-24

solving, cos (x) = 58.3°

However,

When I solve , still using eq (1) , but energy consevation rather than eq. 2 :

Incident photon energy (E) = Electron KE (A) + Scattered photon energy (B)

Then E-B= A

and E-B = hc/Δλ = hc/(2h/mc) = mc^2/2

Then electron KE = γmc^2 - mc^2 = mc^2/2

then : γ= 3/2

and solving γ for v^2, v^2= 5/9(c^2)

Therefore (pe) =mvλ = 3.056*10^-22 ( 4sf)

Finally, subbing this back into eq(1) :

pe*cos(x) = 3.2966*10^-24

3.056*10^-22*cos(x) = 3.2966*10^-24

and so cos (x) = 0.0107...

Any help greatly appreciated !Many , many thanks.

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# Homework Help: Compton Scattering - angle of deflected electron using momentum and energy conservati

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