Compton Scattering - determine momentum

In summary, the initial wavelength gives the total momentum, p=h/11.2p. Which is 59.161y. The problem statement says to "answer in multiples of 10-22 kg.m/s". Is your "y" supposed to equal 10-22 kg.m/s? If so, your 59.161y isn't quite right. Check your arithmetic. Then I tried to substract the momentum from the scattered light to get the momentum of the electron. Vector arithmetic wasn't enough, so I needed to use vector diagramming to find the answer. I found that the x-component of the electron momentum was 0.
  • #1
JoeyBob
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Homework Statement
see attached
Relevant Equations
p = h/wavelenghth
So the initial wavelength gives the total momentum, p=h/11.2p. Which is 59.161y.

Then I tried to substract the momentum from the scattered light to get the momentum of the electron.

59.161y-h/13.6p, which ends up being 0.4872 as the final answer, but the answer is supposed to be 0.77?
 

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  • #2
JoeyBob said:
Homework Statement:: see attached
Relevant Equations:: p = h/wavelenghth

So the initial wavelength gives the total momentum, p=h/11.2p. Which is 59.161y.

Then I tried to substract the momentum from the scattered light to get the momentum of the electron.

59.161y-h/13.6p, which ends up being 0.4872 as the final answer, but the answer is supposed to be 0.77?
Here are a few questions to think about:
1. Are 'p' and 'y' meant to be units?!
2. What do you know about the directions of the electron and the scattered photon after the collision (e.g. include a diagram)?
3. What equation relates the wavelengths of the incident and scattered photons to the the photon scattering angle?
4. Do you know how to add/subtract vectors which have different directions?
 
  • #3
JoeyBob said:
So the initial wavelength gives the total momentum, p=h/11.2p. Which is 59.161y.
The problem statement says to "answer in multiples of 10-22 kg.m/s". Is your "y" supposed to equal 10-22 kg.m/s? If so, your 59.161y isn't quite right. Check your arithmetic.
JoeyBob said:
Then I tried to substract the momentum from the scattered light to get the momentum of the electron.
The momenta of the incoming and outgoing photons are vectors. They're not in the same direction, so you can't simply subtract the magnitudes. You need to use vector arithmetic.

Hint: draw a vector diagram. You should find that the arithmetic is actually rather simple.
 
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  • #4
jtbell said:
The problem statement says to "answer in multiples of 10-22 kg.m/s". Is your "y" supposed to equal 10-22 kg.m/s? If so, your 59.161y isn't quite right. Check your arithmetic.

The momenta of the incoming and outgoing photons are vectors. They're not in the same direction, so you can't simply subtract the magnitudes. You need to use vector arithmetic.

Hint: draw a vector diagram. You should find that the arithmetic is actually rather simple.

But the question doesn't give vectors, so how am I suppose to draw a vector diagram or assign vectors if none are known?

y=yocto or 10-24.
 
  • #5
JoeyBob said:
But the question doesn't give vectors, so how am I suppose to draw a vector diagram or assign vectors if none are known?

y=yocto or 10-24.
You can easily look up the diagram/standard formula for Compton scattering. These should already be in your notes/textbook if you have been taught Compton scattering.
E.g. http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compton.html

Can you now see how to proceed?

Also note, it is incorrect to use a prefix as a unit. For example for a picometre you should write 'pm' not just 'p'.
 
  • #6
JoeyBob said:
y=yocto or 10-24.
OK, so your number for the incoming photon momentum agrees with mine, which is 5.9161 x 10-23 kg.m/s, or 0.59161 x 10-22 kg.m/s using the terms of the problem statement.
JoeyBob said:
how am I suppose to draw a vector diagram or assign vectors if none are known?
You can choose the incoming photon momentum (total momentum of the system) to be in the x-direction. The standard Compton-scattering formula gives you the direction (angle) of the outgoing photon momentum. This is enough information to calculate the outgoing electron momentum, both magnitude (which is what you want) and direction (maybe you can get bonus credit for that :wink: ).
 
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  • #7
jtbell said:
OK, so your number for the incoming photon momentum agrees with mine, which is 5.9161 x 10-23 kg.m/s, or 0.59161 x 10-22 kg.m/s using the terms of the problem statement.

You can choose the incoming photon momentum (total momentum of the system) to be in the x-direction. The standard Compton-scattering formula gives you the direction (angle) of the outgoing photon momentum. This is enough information to calculate the outgoing electron momentum, both magnitude (which is what you want) and direction (maybe you can get bonus credit for that :wink: ).

Ok so I used the formula

change in wavelength = (h/mc)(1-cos(angle) to get an angle of 89.422 degrees, but isn't this giving the angle the photon is scattered and not the angle the electron is scattered?

I could understand how I could use this vector to for lights momentum, but i don't see how I can use it for the electron.
 
  • #8
Hint: Use conservation of momentum. Total initial (incoming) vector momentum = total final (outgoing) vector momentum.
 
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  • #9
so

0.59161 x 10-22 = cos(89.422) * magnitude light momentum + x component of electron momentum

So I really now just need to find the magnitude of light momentum after scattering occurs.

The scattered light wavelength ss 13.6 x 10-12, p=h/wavelength gives 4.8721 x 10-23.

Using the above equation to calculate x component of electron momentum I get a momentum of 0.5867 x 10-22. But it should be 0.77 x 10-22.

Looking into this, the correct answer doesn't even make sense to me unless the light has a negative component, since the x momentum of the electron is greater than the initial x momentum of the light.

But even if i add the final momentum of the light to the initial, I still only get 0.597 x 10-22.

I could be doing something wrong with converting magnitude to x component, multiplying the magnitude by cos(89.422). I haven't used vectors in a long time, was hoping id never have to see them again.

Or I could have calculated the angle incorrectly. Or something else.
 
  • #10
JoeyBob said:
Using the above equation to calculate x component of electron momentum I get a momentum of 0.5867 x 10-22.
OK, that's what I get, too.
JoeyBob said:
But it should be 0.77 x 10-22.
No. that number is what the magnitude of the electron momentum vector (not the x-component) is supposed to turn out to be.

If you know the x-component of a vector, how can you find the magnitude, and what else do you need to know in order to do it? There are actually two ways to do it, but one of them is a bit easier, given the information that you have.
 
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  • #11
Alternate Method:

$$E_k=\frac{hf_1-hf_2}{q_e}\;eV$$
$$\gamma=\frac{E_k+E_0}{E_0}=\sec\delta,$$ where ##\sin\delta=\frac{v}{c}## and ##E_k,E_0## are kinetic energy and rest energy of the electron respectively (in eV).
$$p_e=m_ec\tan\delta$$
 
  • #12
neilparker62 said:
Alternate Method:

$$E_k=\frac{hf_1-hf_2}{q_e}\;eV$$
$$\gamma=\frac{E_k+E_0}{E_0}=\sec\delta,$$ where ##\sin\delta=\frac{v}{c}## and ##E_k,E_0## are kinetic energy and rest energy of the electron respectively (in eV).
$$p_e=m_ec\tan\delta$$
Your first equation has energy on the left but energy/charge on the right.
 
  • #13
Steve4Physics said:
Your first equation has energy on the left but energy/charge on the right.
This gives the energy in electron-volts (eV) instead of joules. I actually prefer to use electron-volt units for energy, mass and momentum, because it eliminates a lot of factors of c in relativistic calculations. However, the problem statement clearly calls for SI units, so I strongly suspect the OP has not used eV-units extensively (if at all!) in relativistic calculations.

Starting with conservation of energy, instead of conservation of momentum, is a valid approach. This avoids having to mess with vectors. If you can find the energy of the outgoing electron, you can then find the magnitude of its momentum.

IMHO the rest of the "alternate method" above might be a bit obscure for someone at the introductory level.
 
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  • #14
jtbell said:
This gives the energy in electron-volts (eV) instead of joules. I actually prefer to use electron-volt units for energy, mass and momentum, because it eliminates a lot of factors of c in relativistic calculations. However, the problem statement clearly calls for SI units, so I strongly suspect the OP has not used eV-units extensively (if at all!) in relativistic calculations.

Starting with conservation of energy, instead of conservation of momentum, is a valid approach. This avoids having to mess with vectors. If you can find the energy of the outgoing electron, you can then find the magnitude of its momentum.

IMHO the rest of the "alternate method" above might be a bit obscure for someone at the introductory level.
Would it perhaps be clearer if I wrote in the last line: $$p_e=m_ec\tan\delta=m_ec\sqrt{\gamma^2-1}$$ ?
 
  • #15
jtbell said:
This gives the energy in electron-volts (eV) instead of joules.
Yes. It just seems unusual to write $$E_k=\frac{hf_1-hf_2}{q_e}\;eV$$ because it embodies a unit in an otherwise purely symbolic equation. As a result, this means ##h,f_1,f_2## and ##q_e## must be in SI units.

I'm just being pedantic I guess.

Edited to correct formatting.
 
  • #16
Indeed, I would write the equation without ##q_e## and units, and use either ##h = 6.626 \times 10^{-34}~\rm{J \cdot s}## or ##h = 4.136 \times 10^{-15}~\rm{eV \cdot s}##, according to my desired units. Textbooks list both values of ##h##, as does Wikipedia. The numerical value of the electron charge here is just a unit conversion factor: ##1~\rm{eV} = 1.602 \times 10^{-19}~\rm{J}##.
 
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  • #17
neilparker62 said:
$$\gamma=\frac{E_k+E_0}{E_0}=\sec\delta,$$ where ##\sin\delta=\frac{v}{c}## and ##E_k,E_0## are kinetic energy and rest energy of the electron respectively (in eV).
$$p_e=m_ec\tan\delta$$
neilparker62 said:
Would it perhaps be clearer if I wrote in the last line: $$p_e=m_ec\tan\delta=m_ec\sqrt{\gamma^2-1}$$ ?
I don't see any reason to introduce the electron's velocity via ##\gamma = 1 / \sqrt{1 - v^2/c^2} = \sec \delta##, when the problem statement doesn't call for finding ##v##. One can use the well-known relativistic relationship among (total = kinetic + rest) energy, momentum and mass: ##E^2 = (pc)^2 + (mc^2)^2##. If one uses eV-units, this is especially convenient because ##E##, ##pc## and ##mc^2## all have units of eV (or keV or MeV, etc.).
 
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  • #18
We don't need v to find ##\gamma## or vv since we already have an expression for ##\gamma## in terms of energy. The only reason for introducing v would be to show that ##p_e=m_ec\tan\delta=m_ec\sqrt{\gamma^2-1}.## Personally I find it convenient to set ##\sin\delta=\frac{v}{c}## when it follows that ##\gamma=\sec\delta##. Then there are convenient trig expressions for electron KE, electron momentum, electron total energy etc. This provides an easy path from electron KE to electron momentum as indicated previously.
 
  • #19
jtbell said:
OK, that's what I get, too.

No. that number is what the magnitude of the electron momentum vector (not the x-component) is supposed to turn out to be.

If you know the x-component of a vector, how can you find the magnitude, and what else do you need to know in order to do it? There are actually two ways to do it, but one of them is a bit easier, given the information that you have.

So either I need to find the angle the electron is ejected or the y component of its momentum. The y component of the photons momentum is sin(89.422)*4.872 x 10-23 = 4.872 x 10-23. The initial y component of the lights momentum is 0. Therefore the y component of the electrons momentum is -4.872 x 10-23.

Using sqrt(x^2+y^2) to give magnitude, I get 7.63 x 10-23, which is only slightly off probably from rounding.

Attempting the energy method, i have the equation Kinetic E of elec. = (hc/wavelength initial) - (hc/wavelength final) in my notes.

So Ek = 1.988 x 10-25((1/11.2 x 10-12) - (1/13.6 x 10-12)) = 3.132 x 10-15.

Ek=mc^2(1/sqrt(1-(v^2/c^2)))-1

Ill spare the details, I get a velocity of 80630000 m/s or 8.063 x 10^7.

Using p=mv/sqrt(1-(v^2/c^2)) I get a momentum of 7.63 x 10-23. Again, off slightly probably from rounding but overall the right answer.

I would say the second way is arguably easier because vectors are bad for you.

Thanks for all the help
 
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  • #20
I got 7.623 x 10-23, three different ways, using an extra couple of digits in c, h, and me, and making sure not to round off intermediate results. Slightly different values for those constants probably explain the difference between us two. I think whoever wrote the problem had a bit of roundoff error.
 
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  • #21
jtbell said:
I got 7.623 x 10-23, three different ways, using an extra couple of digits in c, h, and me, and making sure not to round off intermediate results. Slightly different values for those constants probably explain the difference between us two. I think whoever wrote the problem had a bit of roundoff error.
Agreed - same result per following:

1617983647177.png
 

Related to Compton Scattering - determine momentum

1. What is Compton Scattering and how does it determine momentum?

Compton Scattering is a phenomenon in which a photon collides with a charged particle, resulting in a change in the wavelength and direction of the photon. This change in wavelength can be used to determine the momentum of the charged particle.

2. What is the equation for determining momentum in Compton Scattering?

The equation for determining momentum in Compton Scattering is p = h/λ, where p is the momentum, h is Planck's constant, and λ is the change in wavelength of the photon.

3. How is Compton Scattering used in particle physics?

Compton Scattering is used in particle physics to study the properties of subatomic particles, such as their mass and charge. By measuring the change in wavelength of the scattered photons, scientists can determine the momentum of the particles and use this information to understand their behavior.

4. Can Compton Scattering be used to determine the momentum of all particles?

No, Compton Scattering is only applicable to charged particles. This is because the change in wavelength of the scattered photon is dependent on the charge of the particle it collides with. Neutral particles, such as neutrons, do not produce a change in wavelength and therefore cannot be studied using Compton Scattering.

5. Are there any limitations to using Compton Scattering to determine momentum?

Yes, there are limitations to using Compton Scattering. This technique is most accurate for particles with low energies and small masses. At higher energies, the effects of relativity and quantum mechanics must be taken into account, making the calculations more complex. Additionally, the accuracy of the measurement can be affected by experimental factors such as the angle and energy of the incident photon.

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