- #1

Ale_keys

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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...