# Relationship between spectral distribution and Compton scattering

In summary, the problem involves studying Compton scattering with a distribution of frequencies, specifically a Lorentzian profile. The incident x-rays have a distribution of frequencies, and the problem asks for the calculation of the scattered frequency distribution for each angle ##\theta##. The student is advised to find relevant course material on Lorentzian distribution/profile, and to ask for reference from the course provider. A possible approach suggested by another user is to plot the old Lorentzian profile with new frequencies after Compton scattering to observe the changes.
Homework Statement
You have an x-ray source with photon energy of 17 keV and has a spectral distribution which follows a Lorentzian profile with a full width at half maximum of 0.1 eV.
L=L0/(1+2((w-w0)/delta w)^2)
with w=frequency, w0=central frequency (corresponding to the energy, E), delta w=FWHM,L0=1
Show for the Compton formula showing how the spectrum changes
with the detection angle. Show the output graph for theta = 0-180 degrees with an increment
of 5 degrees.
Relevant Equations
Compton scattering formula: delta lambda=(h/(m_e*c))(1- cos theta)
First of all, this is question from the modern physics module in 1st year physics program. The problem is I have no prior knowledge about spectroscopy or Lorentzian profile. However, the Compton scattering topic was already introduced.

The Compton scattering formula can be changed into the form:
delta(1/f)=(h/m_e*c^2)(1- cos theta)

The w in the homework statement corresponds to omega.

My question is: How is the spectral distribution related to the Compton scattering? Which variable in the Compton scattering formula corresponds to which variable in the Lorentzian profile? Is the spectral distribution produced by the photons emitted by the electrons after Compton scattering?

And also in the question stated omega is the frequency but normally omega is the symbol of angular frequency, so I am not sure that whether omega is frequency or angular frequency.

Thanks.

The problem is I have no prior knowledge about spectroscopy or Lorentzian profile.

My question is: How is the spectral distribution related to the Compton scattering? Which variable in the Compton scattering formula corresponds to which variable in the Lorentzian profile? Is the spectral distribution produced by the photons emitted by the electrons after Compton scattering?

Your first priority is to find the relevant course material on a Lorentzian distribution/profile. Or try online.

You've probably studied Compton scattering for a given frequency/wavelength. This problem is asking you to study it for a distribution of frequencies. The incident x-rays have a distribution of frequency (Lorentzian), hence the x-rays scattered at an angle ##\theta## will also have a distribution of frequency.

For each ##\theta## you are asked to calculate the scattered frequency distribution.

You'll also have to find out what precisely is meant by "output graph". I can't help you there, I'm afraid.

PeroK said:
Your first priority is to find the relevant course material on a Lorentzian distribution/profile. Or try online.

You've probably studied Compton scattering for a given frequency/wavelength. This problem is asking you to study it for a distribution of frequencies. The incident x-rays have a distribution of frequency (Lorentzian), hence the x-rays scattered at an angle ##\theta## will also have a distribution of frequency.

For each ##\theta## you are asked to calculate the scattered frequency distribution.

You'll also have to find out what precisely is meant by "output graph". I can't help you there, I'm afraid.

Do you have any suggestions on course on Lorentzian profile? Fotr now the syllabus I studied only covered until Krane's Chapter 2 to 5 at most.

Do you have any suggestions on course on Lorentzian profile? Fotr now the syllabus I studied only covered until Krane's Chapter 2 to 5 at most.
Sorry, I haven't worked with it. Are you self-studying?

PeroK said:
Sorry, I haven't worked with it. Are you self-studying?
No, I am not self-studying. But Lorentzian profile is never on the syllabus in the module"Modern Physics", since the lectures are almost based on topics from Krane.

No, I am not self-studying. But Lorentzian profile is never on the syllabus in the module"Modern Physics", since the lectures are almost based on topics from Krane.
You need to ask your course provider for an appropriate reference. The Wikipedia page looks quite general, but you may be able to deduce from that precisely what distribution you have in the question.

https://en.wikipedia.org/wiki/Cauchy_distribution

PeroK said:
You need to ask your course provider for an appropriate reference. The Wikipedia page looks quite general, but you may be able to deduce from that precisely what distribution you have in the question.

https://en.wikipedia.org/wiki/Cauchy_distribution
I asked him and he asked me to search the internet myself. Never mind,maybe I will search myself, thanks.

I asked him and he asked me to search the internet myself. Never mind,maybe I will search myself, thanks.
Someone else on here is sure to know a lot more about it. Maybe you'll get a better answer later. Good luck!

Although I've never learned about Lorentzian profile, I'd like to have a guess.
$$L = \frac{1}{1+2(\frac{\omega - 17\,KeV/\hbar}{0.1\,eV/\hbar})^2}$$
This profile is just a function of ##\omega##. First we have a series of ##\omega## range from 0 to ##\infty## and can generate a profile (a function). Then, after Compton scattering ##\omega##'s change to new ones with a one-to-one correspondence (dependent on ##\theta##). Then, maybe what you need to do is to plot the old ##L## with new ##\omega##'s and see what happens.

The question seems to be a computational (not analytical) one? I feel that it is OK to generate figures by MATLAB or something.

pai535 said:
Although I've never learned about Lorentzian profile, I'd like to have a guess.
$$L = \frac{1}{1+2(\frac{\omega - 17\,KeV/\hbar}{0.1\,eV/\hbar})^2}$$
This profile is just a function of ##\omega##. First we have a series of ##\omega## range from 0 to ##\infty## and can generate a profile (a function). Then, after Compton scattering ##\omega##'s change to new ones with a one-to-one correspondence (dependent on ##\theta##). Then, maybe what you need to do is to plot the old ##L## with new ##\omega##'s and see what happens.The question seems to be a computational (not analytical) one? I feel that it is OK to generate figures by MATLAB or something.
My lecturer asked for Python coding, so I think MATLAB cannot be used.
And also, do you mean that:
$$\omega=\omega_i=\frac{1}{\frac{1}{\omega_f}-\frac{h}{m_e c^2}(1-\cos{\theta})}$$

But using Desmos to plot gives me a straight horizontal line at 0. Maybe I will check again...

EDIT: Found the peak but on the frequency is on the magnitude of 10^18, I guess this is why the graph produced is problematic.

Last edited:
Homework Statement:: You have an x-ray source with photon energy of 17 keV and has a spectral distribution which follows a Lorentzian profile with a full width at half maximum of 0.1 eV.
L=L0/(1+2((w-w0)/delta w)^2)
with w=frequency, w0=central frequency (corresponding to the energy, E), delta w=FWHM,L0=1
Show for the Compton formula showing how the spectrum changes
with the detection angle. Show the output graph for theta = 0-180 degrees with an increment
of 5 degrees.
Relevant Equations:: Compton scattering formula: delta lambda=(h/(m_e*c))(1- cos theta)
.
I’m on the limits of what I know, but here are some thoughts.

E = 17keV and ΔE = 0.1eV = 0.0001keV.

Such a narrow line-width seems unlikely. The results would probably be indistinguishable from a monochromatic 17keV source. Are you sure you have values/units correct? Or maybe you are meant to realize this (monochromatic approximation) for yourself?

The question states “Show for the Compton formula showing how the spectrum changes
with the detection angle.” This doesn’t make 100% sense grammatically.

You need to be sure you know what ‘spectrum’ means in this context; it could mean:
a) intensity at different angles;
b) photon energy at different angles.

If b), you would just use your ‘Compton formula’ to find the energy at different angles.

But if a), it’s much harder. The intensity at a given angle corresponds to what is called the differential cross-section at that angle. So it would mean you are being asked to find the differential cross-section (for 17keV Compton scattered photons) at different angles. This requires a special (relativistic) formula called the Klein-Nishina formula. You will have to look this up if required. However the photon energy (17keV) is quite a lot less than the electron rest-energy (511keV) so, to a decent approximation, you could treat this as non-relativistic and the Klein-Nishina formula simplifies (to give Thompson scattering).

Note that the equation$$L=\frac{L₀}{1+2(\frac{ω - ω₀}{Δω})^2}$$ gives all the information about the shape of the source’s spectrum. You can use the formula without knowing anything else about it. (Assuming you need to use it with such a narrow line-width.) The fact that it’s called ‘Lorentzian’ is unimportant/irrelevant. But if you wanted to know, out of interest, what Lorentzian means look here for example : https://en.wikipedia.org/wiki/Spectral_line_shape

Angular frequency (ω), conventional frequency (f) and photon energy (E) are mutually proportional so we can write the above equation with f or E substituting for ω (conversion factors cancel).

With energy in keV and the given values this gives $$L=\frac{1}{(1+2(\frac{E – 17}{0.0001})^2}$$(Less messy than working Hz)

But, as already noted, this is such a narrow line (assuming your values are correct) that you might as well treat the X-rays as monochromatic. This would mean you don't use the above formula.

Steve4Physics said:
I’m on the limits of what I know, but here are some thoughts.

E = 17keV and ΔE = 0.1eV = 0.0001keV.

Such a narrow line-width seems unlikely. The results would probably be indistinguishable from a monochromatic 17keV source. Are you sure you have values/units correct? Or maybe you are meant to realize this (monochromatic approximation) for yourself?

The question states “Show for the Compton formula showing how the spectrum changes
with the detection angle.” This doesn’t make 100% sense grammatically.

You need to be sure you know what ‘spectrum’ means in this context; it could mean:
a) intensity at different angles;
b) photon energy at different angles.

If b), you would just use your ‘Compton formula’ to find the energy at different angles.

But if a), it’s much harder. The intensity at a given angle corresponds to what is called the differential cross-section at that angle. So it would mean you are being asked to find the differential cross-section (for 17keV Compton scattered photons) at different angles. This requires a special (relativistic) formula called the Klein-Nishina formula. You will have to look this up if required. However the photon energy (17keV) is quite a lot less than the electron rest-energy (511keV) so, to a decent approximation, you could treat this as non-relativistic and the Klein-Nishina formula simplifies (to give Thompson scattering).

Note that the equation$$L=\frac{L₀}{1+2(\frac{ω - ω₀}{Δω})^2}$$ gives all the information about the shape of the source’s spectrum. You can use the formula without knowing anything else about it. (Assuming you need to use it with such a narrow line-width.) The fact that it’s called ‘Lorentzian’ is unimportant/irrelevant. But if you wanted to know, out of interest, what Lorentzian means look here for example : https://en.wikipedia.org/wiki/Spectral_line_shape

Angular frequency (ω), conventional frequency (f) and photon energy (E) are mutually proportional so we can write the above equation with f or E substituting for ω (conversion factors cancel).

With energy in keV and the given values this gives $$L=\frac{1}{(1+2(\frac{E – 17}{0.0001})^2}$$(Less messy than working Hz)

But, as already noted, this is such a narrow line (assuming your values are correct) that you might as well treat the X-rays as monochromatic. This would mean you don't use the above formula.

Firstly, the values 17keV and 0.1eV are as in the original question.

For the grammatical problem, the original question is worded as:" Write a code showing for...", with the rest followed word-by-word as in the homework statement.

For the context of the word "spectrum", I don't know either. No any previous context is given before this question. Maybe I will ask my lecturer to clarify.

So, if "spectrum" means photon energy at different angles, does this means that I can substitute the Compton formula into the spectral function and will have the variables ##\omega_i## for initial frequency of photons in Compton scattering and the scattering angle ##\theta## for the new spectral function?

So, if "spectrum" means photon energy at different angles, does this means that I can substitute the Compton formula into the spectral function and will have the variables ##\omega_i## for initial frequency of photons in Compton scattering and the scattering angle ##\theta## for the new spectral function?
No. That won't work. See if this makes sense...

It is not difficult to use the Compton formula to get an expression for the energy of a scattered photon (Eₛ). You end up with$$\frac{1}{E_s} = \frac{1}{E_i} + \frac{1}{m_ec^2}(1 - cosθ)$$where Eᵢ and Eₛ are energies of the incident and scattered photons.

Since the electron’s rest mass is 511keV, working in units of keV gives:$$\frac{1}{E_s} = \frac{1}{17} + \frac{1}{511}(1 - cosθ)$$$$E_s = [{\frac{1}{17} + \frac{1}{511}(1 – cosθ)}]^{-1}$$You can then use this formula to find values of Eₛ at different angles. Then plot Eₛ vs. θ.

We are treating the source as monochromatic (every incident photon is exactly Eᵢ = 17keV). As a result, each angle has a unique value of Eₛ. That’s important to understand.

Now, for the purposes of explanation, let’s suppose the source is not monochromatic but has a wide spread of energy, e.g. ΔE = 3keV.

Some of the incident photons will be Eᵢ=17.9keV, or Eᵢ=20.2keV or Eᵢ=15.2keV for example. That means there is a wide range of possible incident energies and, for a given value of θ, there is a correspoinding wide range of possible scattered energies. We cannot draw a graph of Eₛ vs. θ because there isn’t a unique value of Eₛ for each value of θ.

This means, unless we have a monochromatic source, a graph of Eₛ vs. θ is meaningless; so it cannot be what is meant by ‘spectrum’.

You need to find out:
a) what is meant by ‘spectrum’. The x-axis is angle, but what is the y-axis?
b) is ΔE really 0.1eV? And, if so, can you treat the X-rays as monochromatic?

You can’t really get any further unless you have answers to these questions.

It would also help to know what your have covered. Do you know what a scattering cross-section is? Have you covered Thomson and Kline-Nishima scattering formulae?

Sorry I can’t assist more at this stage. If you can get answers to the above questions, I’ll try to help some more.

Steve4Physics said:
No. That won't work. See if this makes sense...

It is not difficult to use the Compton formula to get an expression for the energy of a scattered photon (Eₛ). You end up with$$\frac{1}{E_s} = \frac{1}{E_i} + \frac{1}{m_ec^2}(1 - cosθ)$$where Eᵢ and Eₛ are energies of the incident and scattered photons.

Since the electron’s rest mass is 511keV, working in units of keV gives:$$\frac{1}{E_s} = \frac{1}{17} + \frac{1}{511}(1 - cosθ)$$$$E_s = [{\frac{1}{17} + \frac{1}{511}(1 – cosθ)}]^{-1}$$You can then use this formula to find values of Eₛ at different angles. Then plot Eₛ vs. θ.

We are treating the source as monochromatic (every incident photon is exactly Eᵢ = 17keV). As a result, each angle has a unique value of Eₛ. That’s important to understand.

Now, for the purposes of explanation, let’s suppose the source is not monochromatic but has a wide spread of energy, e.g. ΔE = 3keV.

Some of the incident photons will be Eᵢ=17.9keV, or Eᵢ=20.2keV or Eᵢ=15.2keV for example. That means there is a wide range of possible incident energies and, for a given value of θ, there is a correspoinding wide range of possible scattered energies. We cannot draw a graph of Eₛ vs. θ because there isn’t a unique value of Eₛ for each value of θ.

This means, unless we have a monochromatic source, a graph of Eₛ vs. θ is meaningless; so it cannot be what is meant by ‘spectrum’.

You need to find out:
a) what is meant by ‘spectrum’. The x-axis is angle, but what is the y-axis?
b) is ΔE really 0.1eV? And, if so, can you treat the X-rays as monochromatic?

You can’t really get any further unless you have answers to these questions.

It would also help to know what your have covered. Do you know what a scattering cross-section is? Have you covered Thomson and Kline-Nishima scattering formulae?

Sorry I can’t assist more at this stage. If you can get answers to the above questions, I’ll try to help some more.
I tried this approach before and get a nice graph with peak at 1 actually, but my coursemates think elsewise, so I decided to ask somewhere for further clarification. Maybe I am right at the beginning...

By the way, maybe a 2 variable graph is expected, by changing ##\theta## using an external slider.(This problem should be solved using coding so this is possible).

For the meaning of "spectrum" I am still waiting the reply from my lecturer.

I am quite sure that the value written on the exercise is 0.1eV, but whether the lecturer ask us to assume that this is a monochromatic source is not known.

I know nothing of the topics you mentioned above. In this introductory module, the class until now only covered Chapter 5 of Krane (very basic SE) with several topics skipped.

Thank you for your assistance until now. I think that maybe any progress can be done only after the reply from my lecturer.

## 1. What is the relationship between spectral distribution and Compton scattering?

The spectral distribution of electromagnetic radiation refers to the intensity of different wavelengths of light in a given source. Compton scattering is a phenomenon in which photons (particles of light) are scattered by free electrons, resulting in a shift in the energy and wavelength of the scattered photons. The spectral distribution of a source can affect the likelihood and extent of Compton scattering, as different wavelengths of light interact with electrons in different ways.

## 2. How does the energy of the incident photon affect Compton scattering?

The energy of the incident photon has a direct impact on the amount of energy transferred to the electron during Compton scattering. As the energy of the photon increases, so does the energy transferred to the electron, resulting in a larger shift in the wavelength of the scattered photon. This relationship is described by the Compton scattering equation: Δλ = h/mc (1-cosθ), where h is Planck's constant, m is the mass of the electron, c is the speed of light, and θ is the angle of scattering.

## 3. How does the angle of scattering affect the spectral distribution of the scattered photons?

The angle of scattering, θ, affects the spectral distribution of the scattered photons by determining the amount of energy transferred to the electron. As the angle of scattering increases, the amount of energy transferred also increases, resulting in a larger shift in the wavelength of the scattered photon. This can be seen in the Compton scattering equation, where the term (1-cosθ) represents the change in energy.

## 4. What is the difference between coherent and incoherent Compton scattering?

Coherent Compton scattering occurs when the incident photon interacts with a bound electron, resulting in a change in the electron's energy level and a shift in the wavelength of the scattered photon. Incoherent Compton scattering, on the other hand, occurs when the incident photon interacts with a free electron, resulting in a change in the electron's kinetic energy and a shift in the wavelength of the scattered photon. Coherent Compton scattering is more likely to occur in low energy photons, while incoherent Compton scattering is more common in high energy photons.

## 5. How does the material of the target affect Compton scattering?

The material of the target can affect Compton scattering in several ways. The number of free electrons in the material can impact the likelihood of incoherent Compton scattering, as more free electrons increase the chances of interaction with the incident photons. Additionally, the atomic number of the material can affect the energy of the scattered photons, as higher atomic numbers result in a larger energy transfer during Compton scattering. Finally, the density of the material can also impact the likelihood of Compton scattering, as denser materials have a higher concentration of free electrons and therefore a higher chance of interaction with the incident photons.

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