What was the angle of deflection?

[Ale_keys]

Homework Statement
The scientist changes the frequency of the incident X-ray to 4.50 x 10^19 Hz and measures the deflected X ray frequency of 4.32 x 10^19 Hz. What was the angle of deflection?

Fi = 4.50 x 10^19 Hz
Ff = 4.32 x 10^19 Hz

2. Homework Equations
Δλ = λf - λi
Δλ = (h/mc)(1-cosθ)
λ = c/f

3. The Attempt at a Solution

First, I found the wavelength of the X ray before and after it is deflected.

λi = c/f
= (3.00 x 10^8 m/s) / (4.50 x 10^19 Hz)
= 6.818181812 x 10^-12 m

λf = c/f
= (3.00 x 10^8 m/s) / (4.32 x 10^19 Hz)
= 6.9444444444 x 10^-12 m

Then found the change in wavelength

Δλ = λf - λi
= (6.944444444 x 10^-12 m) - (6.818181812 x 10^-12 m)
= 1.26262626 x 10^-13 m

Then I used the change in wavelength to find the angle

Δλ = (h/mc)(1-cosθ)
1.26262626 x 10^-13 m = ((6.63 x 10^-34 Js) / (9.11 x 10^-31 kg)(3.00 x 10^8 m/s))(1-cosθ)
1.26262626 x 10^-13 m = (2.4259056 x 10^-12)(1-cosθ)
0.052047623 = 1-cosθ
cosθ = 0.947952377
cos^-1(0.947952377) = 18.56692499° = 18.6°

I'm just really unsure of the process that I took...

 

[Simon Bridge]

Welcome to PF;
Please do not leave out vital information - I am guessing that this is a Compton effect problem.

The way to be more sure of the processes you used, is to write down your reasoning for each step: why did you choose that particular equation?
You should also consider if the answer makes sense in terms of the physics - i.e. if you already know the deflection angle for another frequency, you can compare the deflection angles ... how does the deflection angle generally depend on the frequency: i.e. does a higher frequency produce a bigger or a smaller angle?