X-rays with an energy of 300 keV undergo Compton scattering from a target.
If the scattered x-ray are directed at 30 degrees relative to the incident rays, find
a)The Compton shift at this angle
b) The energy of the scattered x-ray
c)The energy of the recoiling electron
Nothing
Superb
To find the Compton shift, energy of the scattered x-ray, and energy of the recoiling electron, we need to use the Compton scattering equation:
λ' - λ = (h / m_e * c) * (1 - cosθ),
where:
- λ' is the wavelength of the scattered x-ray,
- λ is the wavelength of the incident x-ray,
- h is the Planck constant (6.626 x 10^-34 J·s),
- m_e is the mass of the electron (9.109 x 10^-31 kg),
- c is the speed of light (3.00 x 10^8 m/s),
- θ is the scattering angle.
Given:
- Energy of the incident x-ray = 300 keV,
- Energy of the incident x-ray can be converted to wavelength using λ = hc / E,
- Scattering angle = 30 degrees.
Let's calculate the Compton shift, energy of the scattered x-ray, and energy of the recoiling electron step by step:
a) Compton shift:
The Compton shift is the change in wavelength of the scattered x-ray compared to the incident x-ray.
Step 1: Convert the incident x-ray energy to wavelength:
E = 300 keV = 300 * 10^3 * 1.602 x 10^-19 J (since 1 eV = 1.602 x 10^-19 J)
λ = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (300 * 10^3 * 1.602 x 10^-19 J)
λ = 2.479 x 10^-12 m
Step 2: Calculate the Compton shift using the Compton scattering equation:
θ = 30 degrees = 30 * (π / 180) radians
Compton shift = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (9.109 x 10^-31 kg * 3.00 x 10^8 m/s) * (1 - cos(30 * (π / 180)))
Compton shift ≈ 2.43 x 10^-12 m
Therefore, the Compton shift at this angle is approximately 2.43 x 10^-12 meters.
b) Energy of the scattered x-ray:
The energy of the scattered x-ray can be obtained from the scattered wavelength using the relationship E = hc / λ'.
Step 1: Calculate the wavelength of the scattered x-ray:
λ' = λ + Compton shift
λ' = 2.479 x 10^-12 m + 2.43 x 10^-12 m
λ' ≈ 4.909 x 10^-12 m
Step 2: Calculate the energy of the scattered x-ray using the wavelength:
E' = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (4.909 x 10^-12 m)
E' ≈ 1.274 x 10^-15 J
Therefore, the energy of the scattered x-ray is approximately 1.274 x 10^-15 Joules.
c) Energy of the recoiling electron:
The energy of the recoiling electron can be obtained from the energy conservation equation: E + E' = E_e + E'_e,
where:
- E is the energy of the incident x-ray,
- E' is the energy of the scattered x-ray,
- E_e is the rest mass energy of the electron,
- E'_e is the kinetic energy of the recoiling electron.
Step 1: Convert the energy of the incident x-ray to joules:
E = 300 keV = 300 * 10^3 * 1.602 x 10^-19 J (since 1 eV = 1.602 x 10^-19 J)
Step 2: Calculate the rest mass energy of the electron:
E_e = m_e * c^2
E_e ≈ (9.109 x 10^-31 kg) * (3.00 x 10^8 m/s)^2
E_e ≈ 8.187 x 10^-14 J
Step 3: Calculate the energy of the recoiling electron using the energy conservation equation:
E_e + E'_e = E + E'
E'_e = (E + E') - E_e
E'_e ≈ (300 * 10^3 * 1.602 x 10^-19 J + 1.274 x 10^-15 J) - 8.187 x 10^-14 J
Therefore, the energy of the recoiling electron is approximately (3.055 x 10^-18 J) - (8.187 x 10^-14 J).