The diagram below shows two successive scatterings at locations A and B. In the
ID: 1875262 • Letter: T
Question
The diagram below shows two successive scatterings at locations A and B. In the first scattering, a photon of wavelength 1 hits a free electron at A resulting in a photon of wavelength 2 that moves at an angle 1 to the incident direction. The second photon hits an electron at B and scatters at an angle 2 to the direction of the second photon resulting in a photon of wavelength 3. Determine the angle 1 (in degrees) if 2 = 67.0° and the total shift in wavelength of the combined scattering is 3.83 103 nm
The diagram below shows two successive scatter ngs at locations A and . In the first scattering a photon of wave en th . htsa ee electro at A result ng a photo of wa elength 2 that m es at an angle ,to he incident direction. The second photon hits an electron at B and scatters at an angle e2 to the direction of the second photon resulting in a photon of wavelength , Determine the angle(in degrees ife,-67.0° and the total shift in wavelength of the combined scattering is 3.83 x 103 nm. e1 o relative to the incident direction Submit Answer Save Progress Pactice Another VersionExplanation / Answer
from the given data
after first collision let the wavelength of light become lambda'
then from compton scattering
lambda' - lambda = (h/mc)(1 - cos(theta1))
and after second scattering, let the wavelength be lambda"
lambda" - lambda' = (h/mc)(1 - cos(theta2))
adding both
lambda" - lambda = (h/mc)(2 - cos(theta1) - cos(theta2))
now from the given data
lambda" - lambda = 3.83*10^-3 nm
m = 9.1*10^-31 kg
c = 3*10^8 m/s
theta1 = ?
theta2 = 67 deg
hence
theta1 = 88.154 deg
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