Gamma rays of photon energy 0.511 MeV are directed onto an aluminum target and are scattered in various directions by loosely bound electrons there.a) What is the wavelength of the incident gamma rays?
b) What is the wavelength of the gamma rays scattered at 90 degrees to the incident beam?
c) What is the photon energy of the rays scattered in this direction?
To answer these questions, we need to use the principles of Compton scattering. Compton scattering occurs when photons interact with electrons and change their direction, resulting in a shift in their wavelength. The shift in wavelength can be determined by the change in scattering angle.
Let's use the following formulas:
a) The energy of a photon is given by the equation: E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.0 x 10^8 m/s), and λ is the wavelength.
b) For Compton scattering, we use the Compton wavelength shift formula: Δλ = λ' - λ = h/mc * (1 - cosθ), where Δλ is the change in wavelength, λ' is the scattered wavelength, θ is the scattering angle, and m is the rest mass of the electron (9.11 x 10^-31 kg).
c) The energy of a photon can be determined using the same equation as in part a.
Now, let's calculate the values:
a) To find the wavelength of the incident gamma rays, we need to convert the given energy into joules: 0.511 MeV = 0.511 x 10^6 eV = 0.511 x 10^6 x 1.6 x 10^-19 J = 8.176 x 10^-14 J.
Using the formula E = hc/λ, we can rearrange the equation to solve for λ:
λ = hc/E.
Substituting the values, we get:
λ = (6.626 x 10^-34 J·s * 3.0 x 10^8 m/s) / (8.176 x 10^-14 J) ≈ 2.41 x 10^-12 m.
Therefore, the wavelength of the incident gamma rays is approximately 2.41 picometers (pm).
b) For the wavelength of the gamma rays scattered at 90 degrees to the incident beam, we can use the Compton wavelength shift formula.
Let's assume the scattered wavelength is λ' and the scattering angle is θ = 90 degrees.
Δλ = λ' - λ = h/mc * (1 - cosθ).
We know the values of h, m, and c.
Substituting these values, we get:
(λ' - λ) = (6.626 x 10^-34 J·s / (9.11 x 10^-31 kg * 3.0 x 10^8 m/s)) * (1 - cos(90°)).
Simplifying, we find:
(λ' - λ) = 2.43 x 10^-12 m.
Therefore, the wavelength of the gamma rays scattered at 90 degrees to the incident beam is approximately 2.43 picometers (pm).
c) To find the photon energy of the rays scattered in this direction, we can use the same equation as in part a: E = hc/λ.
Substituting the wavelength (λ = 2.43 x 10^-12 m) into the equation, we get:
E = (6.626 x 10^-34 J·s * 3.0 x 10^8 m/s) / (2.43 x 10^-12 m) ≈ 81.5 keV.
Therefore, the photon energy of the rays scattered at 90 degrees to the incident beam is approximately 81.5 kiloelectron volts (keV).
To solve these questions, we can utilize the following equation:
λ = hc/E
where:
λ is the wavelength of the photon,
h is Planck's constant (6.626 x 10^-34 J·s),
c is the speed of light (3 x 10^8 m/s),
E is the energy of the photon.
Given the energy of the incident gamma rays as 0.511 MeV, we can convert it to joules:
E = 0.511 MeV = 0.511 x 10^6 eV = 0.511 x 10^6 x 1.6 x 10^-19 J = 8.176 x 10^-14 J
Now we can calculate the answers to each part of the question:
a) Wavelength of the incident gamma rays:
λ = hc/E
λ = (6.626 x 10^-34 J·s) x (3 x 10^8 m/s) / (8.176 x 10^-14 J)
λ ≈ 2.418 x 10^-12 m
b) Wavelength of the gamma rays scattered at 90 degrees:
The wavelength is not affected by the scattering angle in this case. Hence, it will remain the same as the incident gamma rays:
λ ≈ 2.418 x 10^-12 m
c) Photon energy of the rays scattered at 90 degrees:
The photon energy is given by the equation E = hc/λ.
Using the wavelength from part b:
E = (6.626 x 10^-34 J·s) x (3 x 10^8 m/s) / (2.418 x 10^-12 m)
E ≈ 8.168 x 10^-14 J = 0.511 MeV
So, the photon energy of the gamma rays scattered at 90 degrees is also approximately 0.511 MeV.