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.

i found the photon energy using compton scatter equation at 110 degrees, but can't find the rest.

30 keV x-ray photon strikes the electron initially at rest and the photon is scattered through an angle of 300, what is the recoil velocity of electron?

To find the energy of the photon scattered at 110 degrees, you can use the Compton scattering equation:

λ' - λ = (h / m_e * c) * (1 - cosθ)

Where:
λ' = Wavelength of the scattered photon
λ = Wavelength of the incident photon
h = Planck's constant
m_e = Mass of the electron
c = Speed of light
θ = Scattering angle (110 degrees)

Since you already have the energy of the incident photon (700KeV), you can convert it to wavelength using the equation:

λ = (c / E) * 10^9

where E is the energy of the incident photon in electron volts (700KeV = 700,000eV).

Substituting the values into the Compton scattering equation:

λ' - λ = (h / m_e * c) * (1 - cosθ)

Substituting the value for λ and rearranging the equation, we get:

λ' = λ + (h / m_e * c) * (1 - cosθ)

Now calculate the wavelength (λ):

λ = (c / E) * 10^9
= (3 x 10^8 m/s) / (700,000eV) * 10^9
= 4.28571 x 10^-3 nm

Substituting the values into the equation for λ', we get:

λ' = 4.28571 x 10^-3 nm + (6.63 x 10^-34 Js / (9.11 x 10^-31 kg * 3 x 10^8 m/s)) * (1 - cos(110 degrees))

Now, you can convert the wavelength back to energy using the equation:

E' = (c / λ') * 10^-9

Calculate the energy (E') of the scattered photon using the new wavelength (λ'), and you will find the answer to the first part of your question.

To find the energy of the scattered electron, we can use the conservation of energy and momentum in Compton scattering:

E' + E_e = E + m_e * c^2 (equation 1)
p' + p_e = p (equation 2)

Where:
E' = Energy of the scattered photon (unknown)
E_e = Energy of the scattered electron (unknown)
E = Energy of the incident photon (700KeV in this case)
m_e = Mass of the electron
c = Speed of light
p' = Momentum of the scattered photon (= E' / c)
p_e = Momentum of the scattered electron (unknown)
p = Momentum of the incident photon (= E / c)

From equation 2, we can substitute p' and p with their corresponding values:

(E' / c) + p_e = (E / c)

Rearranging equation 2, we get:

p_e = (E / c) - (E' / c)

Substituting the values into the equation:

p_e = ((E / c) - (E' / c)) * c
= E - E'

Now, you can calculate the energy (E_e) of the scattered electron by subtracting E' from the initial energy (E). This will give you the answer to the second part of your question.

To find the recoil angle of the electron, you can use the equation of the conservation of momentum in the x-direction:

p' * sin(θ) = p_e * sin(φ)

Where:
θ = Scattering angle (110 degrees)
φ = Recoil angle of the electron (unknown)

Substituting the values into the equation, we get:

(E' / c) * sin(θ) = E_e * sin(φ) / c

Rearranging the equation and solving for φ, we get:

φ = asin((E' / E_e) * (sin(θ) * c))

Now, substitute the values into the equation to calculate the recoil angle (φ) of the electron, and you will have the answer to the third part of your question.