If a 6.0keV photon scatters from a fee proton at rest, what is the change in the photon's wavelength if the photon recoils at 90 degrees?
To determine the change in the photon's wavelength after it scatters from a free proton at rest, we need to use the principles of Compton scattering.
Compton scattering is a phenomenon where a photon interacts with a charged particle (such as an electron or a proton) and transfers some of its energy and momentum to the particle. As a result of this transfer, the photon's wavelength increases, which is known as the Compton shift.
The Compton shift can be calculated using the following equation:
Δλ = λ' - λ = h / (m₀c) * (1 - cosθ)
Where:
Δλ is the change in wavelength
λ' is the wavelength after scattering
λ is the initial wavelength
h is the Planck's constant (6.626 x 10^-34 J·s)
m₀ is the rest mass of the proton (1.67 x 10^-27 kg)
c is the speed of light (3.00 x 10^8 m/s)
θ is the scattering angle (90 degrees in this case)
Let's plug in the values and calculate the change in wavelength:
Δλ = (6.626 x 10^-34 J·s) / [(1.67 x 10^-27 kg) * (3.00 x 10^8 m/s)] * (1 - cos(90°))
Now, let's simplify the equation:
Δλ = 2.41 x 10^-12 m
Therefore, the change in the photon's wavelength after scattering from a free proton at rest at a 90-degree angle is approximately 2.41 x 10^-12 meters.