Finding the Scattering Angle of a Photon

[binbagsss]

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.

 

[Simon Bridge]

E-B = hc/Δλ

Why not:
$$\frac{hc}{\lambda^\prime}-\frac{hc}{\lambda}$$

 

[binbagsss]

that is genius.
haha thank you :)

 

[Simon Bridge]

No worries - it's easier to spot these things in someone elses work than in your own.