A gamma ray with 700KeV energy is Compton-Scattered from an electron. Find the energy of the photon scattered at 110 degrees, the energy of the scattered electron, and the recoil angle of the electron.
work out
To find the energy of the photon scattered at 110 degrees, the energy of the scattered electron, and the recoil angle of the electron, we need to use the Compton scattering formula, which relates the initial and final energies and scattering angles of the gamma ray and electron.
The formula for Compton scattering is:
λ' - λ = (h / m_ec) * (1 - cosθ)
Where:
- λ' is the wavelength of the scattered photon
- λ is the wavelength of the incident photon
- h is the Planck's constant (6.626 x 10^-34 Js)
- m_e is the mass of the electron (9.11 x 10^-31 kg)
- c is the speed of light (3 x 10^8 m/s)
- θ is the scattering angle
First, we need to convert the energy of the gamma ray photon to its corresponding wavelength using the energy-wavelength relation:
E = h * c / λ
Rearranging the formula, we have:
λ = h * c / E
Let's calculate the initial wavelength (λ) and the wavelength of the scattered photon (λ') using the given energy of the gamma ray photon (700 KeV).
Initial energy of gamma ray photon (E) = 700 KeV = 700 x 10^3 eV
h = 6.626 x 10^-34 Js
c = 3 x 10^8 m/s
Using the energy-wavelength relationship:
λ = h * c / E = (6.626 x 10^-34 Js) * (3 x 10^8 m/s) / (700 x 10^3 eV * 1.6 x 10^-19 J/eV)
λ ≈ 0.026 nm
Next, we can use the Compton scattering formula to find the energy of the scattered photon and the energy of the scattered electron.
For the energy of the scattered photon:
Given:
θ = 110 degrees
Substituting the values into the Compton scattering formula:
λ' - λ = (h / m_e * c) * (1 - cosθ)
λ' - λ = (6.626 x 10^-34 Js) / (9.11 x 10^-31 kg * 3 x 10^8 m/s) * (1 - cos(110))
λ' ≈ 0.026 nm + 0.047 nm
λ' ≈ 0.073 nm
Now, we can find the energy of the scattered photon by using the energy-wavelength relationship:
E' = h * c / λ'
E' = (6.626 x 10^-34 Js) * (3 x 10^8 m/s) / (0.073 nm * 1 x 10^-9 m/nm)
E' ≈ 270 KeV
Therefore, the energy of the photon scattered at 110 degrees is approximately 270 KeV.
To calculate the energy of the scattered electron, we need to use the energy-momentum conservation principle. Since the energy of the incident photon is known (700 KeV), the energy of the scattered photon (270 KeV), and the mass of the electron (m_e), we can calculate the energy of the scattered electron using:
Energy of scattered electron = Energy of incident photon - Energy of scattered photon
Energy of scattered electron = 700 KeV - 270 KeV
Energy of scattered electron = 430 KeV
Finally, to find the recoil angle of the electron, we can use the momentum conservation principle. The momentum of the scattered electron can be equal to the initial momentum of the electron (which is zero) plus the momentum transferred to the electron by the scattered photon:
p_initial + p_photon = p_scattered_electron
Since the magnitude of the momentum is given by p = √(2m_e*E), we have:
0 + √(2m_e*E') * cos(θ_recoil) = √(2m_e*E)
Rearranging the equation and solving for the recoil angle (θ_recoil):
cos(θ_recoil) = √(2E / E') * cos(θ)
θ_recoil = arccos(√(2E / E') * cos(θ))
Substituting the known values:
θ_recoil = arccos(√(2 * (9.11 x 10^-31 kg) * (430 x 10^3 eV * 1.6 x 10^-19 J/eV) / (9.11 x 10^-31 kg * (700 x 10^3 eV * 1.6 x 10^-19 J/eV))) * cos(110))
θ_recoil ≈ 133 degrees
Therefore, the energy of the photon scattered at 110 degrees is approximately 270 KeV, the energy of the scattered electron is 430 KeV, and the recoil angle of the electron is approximately 133 degrees.