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
mp=mass of proton=1.673x10^-27
λ=h/mpc(1-cosθ)
λ=(6.6261x10^-34/(1.673x10^-27*3x10^8))(1-cos110)
λ=1.771^-15m
f=c/λ
f=3x10^8/1.771x10^-15
f=1.693x10^23 Hz
h=4.1357x10^-15eV/Hz
E=hf
E=(4.1357x10^-15*1.693x10^23)
E=7.0x10^8eV
To find the energy of the photon scattered at 110 degrees, we can use the equation E = hf, where E is the energy, h is Planck's constant (4.1357x10^-15 eV/Hz), and f is the frequency.
First, let's find the frequency of the scattered photon. We can use the equation f = c/λ, where f is the frequency, c is the speed of light (3x10^8 m/s), and λ is the wavelength.
Given that λ = 1.771x10^-15 m (from the previous calculations), we can calculate f as follows:
f = 3x10^8 m/s / 1.771x10^-15 m
f ≈ 1.693x10^23 Hz
Now we can calculate the energy of the scattered photon:
E = h*f
E = 4.1357x10^-15 eV/Hz * 1.693x10^23 Hz
E ≈ 7.0x10^8 eV
Therefore, the energy of the photon scattered at 110 degrees is approximately 7.0x10^8 eV.
To find the energy of the scattered electron, we can use the conservation of energy and momentum in Compton scattering. The energy of the scattered electron can be approximated using the equation:
E' = E - E'
Where E is the energy of the incident photon (700 KeV = 7.0x10^5 eV) and E' is the energy of the scattered photon (calculated earlier as 7.0x10^8 eV). Rearranging the equation, we get:
E' = E / (1 + E/mc^2)
Substituting the values:
E' = (7.0x10^5 eV) / (1 + (7.0x10^5 eV) / (9.11x10^-31 kg * (3x10^8 m/s)^2))
Using the given mass of the electron (9.11x10^-31 kg), we can calculate the energy of the scattered electron.
Finally, to find the recoil angle of the electron, we can use the equation for the Compton scattering angle:
cos(theta) = 1 + (E/mc^2) / (E'/mc^2)
Solve for theta to find the recoil angle.
Note: It's important to note that these calculations involve many approximations and may not be exact due to rounding errors or other factors. It's always good to double-check the calculations and consult reliable sources for more accurate values.
To find the energy of the photon scattered at 110 degrees, we can use the equation E = hf, where E is the energy of the photon and f is the frequency of the photon.
First, let's calculate the frequency of the scattered photon. We have already determined the wavelength (λ) using the formula λ = h / (mpc * (1 - cosθ)), where h is Planck's constant, mp is the mass of a proton, c is the speed of light, and θ is the scattering angle.
Substituting the given values, we have:
λ = (6.6261 x 10^-34 J*s) / (1.673 x 10^-27 kg * 3 x 10^8 m/s * (1 - cos(110°)))
Calculating this gives us a wavelength of λ = 1.771 x 10^-15 m.
Next, we can calculate the frequency (f) using the equation f = c / λ, where c is the speed of light.
f = (3 x 10^8 m/s) / (1.771 x 10^-15 m)
Calculating this gives us a frequency of f = 1.693 x 10^23 Hz.
Finally, we can calculate the energy (E) using the equation E = hf, where h is Planck's constant.
E = (6.6261 x 10^-34 J*s) * (1.693 x 10^23 Hz)
E ≈ 1.119 x 10^-10 J
To convert this energy to electronvolts (eV), we can use the conversion factor 1 eV = 1.6 x 10^-19 J.
E ≈ (1.119 x 10^-10 J) / (1.6 x 10^-19 J/eV)
E ≈ 699.375 eV
So, the energy of the scattered photon is approximately 699.375 eV.