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.

 

[asteriatic]

Hello, it seems like you have already done a great job in your calculations. However, I would like to point out a few things that might help you in solving the problem.

Firstly, when solving for the angle x, you have used the momentum conservation equation (eq. 1) which is correct. However, in your calculation, you have used the momentum of the scattered photon (q) as 3.2966*10^-24, which is actually the momentum of the incident photon (p). The momentum of the scattered photon can be calculated using the equation p = h/λ - h/λ', where λ' is the wavelength after scattering. So, q = h/λ - h/λ' = 3.2966*10^-24.

Secondly, when solving using energy conservation, you have correctly calculated the electron kinetic energy as mc^2/2. However, when solving for the velocity of the electron, you have used the formula v^2 = 5/9(c^2), which is incorrect. The correct formula is v^2 = (5/9)(c^2) * (1 - (mc^2/E)^2). Using this formula, you can calculate the velocity of the electron as v = 0.547c.

Lastly, when you substitute the correct values for the momentum (pe) and velocity (v) in eq. 1, you will get cos(x) = 0.99999, which is essentially equal to 1. This means that the angle x is very close to 0 degrees, which makes sense as the electron is scattered in the same direction as the incident photon.

I hope this helps and clarifies your calculations. Keep up the good work!