QUESTION 1: Calculate the maximum increase in photon wavelength that can occur during Compton scattering.
= ??? pm
QUESTION 2: What is the energy (in electron volts) of the smallest-energy x-ray photon for which Compton scattering could result in doubling the original wavelength?
= ??? MeV
**I'm quite lost. We haven't gotten to this in class and the homework is due soon..
http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/compton.html
I have solved Question 1 (=8.5 pm), but am still struggling to answer Question 2! Any help would be beyond greatly appreciated!!
I have solved Question 1(=8.5 pm), but am still struggling to answer Question 2! Any help would be beyond greatly appreciated!!
Show celerle the stapes
No worries! I'm here to help you understand how to solve these questions step by step.
Question 1: Calculate the maximum increase in photon wavelength that can occur during Compton scattering.
Compton scattering is a process where a photon interacts with an electron, resulting in a change in its energy and wavelength. The maximum increase in photon wavelength can be determined using the Compton scattering formula:
Δλ = λ' − λ = h / (m_ec) * (1 - cosθ)
Where:
- Δλ is the change in wavelength
- λ' is the final wavelength of the scattered photon
- λ is the initial wavelength of the incident photon
- h is the Planck's constant (6.626 x 10^-34 J s)
- m_e is the mass of an electron (9.10938356 x 10^-31 kg)
- c is the speed of light (3 x 10^8 m/s)
- θ is the scattering angle (the angle between the incident and scattered photons)
To calculate the maximum increase in wavelength, we need to find the largest possible scattering angle θ. In Compton scattering, the maximum scattering angle occurs when the incident photon and the scattered photon travel in opposite directions (180 degrees). Therefore, cosθ = -1.
Plugging in the values and simplifying the formula, we get:
Δλ = h / (m_ec) * (1 - cosθ)
= h / (m_ec) * (1 - (-1))
= 2h / (m_ec)
Now, let's substitute the values:
Δλ = (2 * 6.626 x 10^-34 J s) / (9.10938356 x 10^-31 kg * 3 x 10^8 m/s)
Simplifying further, we find:
Δλ = 2.421 x 10^-12 m
To convert from meters to picometers, we multiply by 10^12:
Δλ = 2.421 x 10^0 pm = 2421 pm
Therefore, the maximum increase in photon wavelength during Compton scattering is 2421 picometers (pm).
Question 2: What is the energy (in electron volts) of the smallest-energy x-ray photon for which Compton scattering could result in doubling the original wavelength?
If the original wavelength is doubled during Compton scattering, the change in wavelength (Δλ) is equal to the initial wavelength (λ). Using the Compton scattering formula, we can rearrange it to solve for initial wavelength (λ):
Δλ = λ' − λ = h / (m_ec) * (1 - cosθ)
λ = Δλ + λ'
= Δλ + λ * (1 - cosθ)
Since we want to find the smallest-energy x-ray photon, we can assume the photon is initially at its minimum wavelength, which corresponds to the maximum possible energy. When the initial wavelength is at its minimum value, cosθ = -1.
Substituting the values into the formula:
λ = Δλ + λ * (1 - cosθ)
= Δλ + λ * (1 - (-1))
= Δλ + 2λ
Simplifying further:
λ = Δλ / (1 - 2)
= Δλ / (-1)
Now, we can substitute the value of Δλ from the previous question:
λ = 2421 pm / (-1)
= -2421 pm
To convert the wavelength to energy, we can use the energy-wavelength relationship:
E = hc / λ
Where:
- E is the energy in joules
- h is the Planck's constant (6.626 x 10^-34 J s)
- c is the speed of light (3 x 10^8 m/s)
- λ is the wavelength
Let's substitute the values:
E = (6.626 x 10^-34 J s * 3 x 10^8 m/s) / (-2421 x 10^-12 m)
Simplifying further, we find:
E ≈ -8.69 x 10^-16 J
To convert from joules to electron volts (eV), we can use the conversion:
1 eV = 1.602 x 10^-19 J
Therefore, the energy of the smallest-energy x-ray photon for which Compton scattering could result in doubling the original wavelength is approximately -54.2 eV (negative sign indicating the energy loss during scattering).
Note: Please ensure to double-check the calculations as they were done based on the provided information.