A scientist changes the frequency of an incident xray to 4.50 x 10 to the power of 19 Hz and measures the deflected xray frequency of 4.32 x 10 to the power of 19 Hz. What was the angle of deflection
2 years ago
Answered By Harsh P
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)
2 years ago
Answered By Harsh P
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°