Question

Help :( I have been stuck on this problem and can't seem to get it:
A photon with wavelength λ = 0.1440nm is incident on an electron that is initially at rest. If the photon scatters at an angle of 60.0 ∘ from its original direction, what is the magnitude of the linear momentum of the electron just after the collision with the photon?What is the direction of the linear momentum of the electron just after the collision with the photon?
I am using the p=h/lambda and the comptant scatting equation and not getting the right results. Is it correct that angle=tan-1(p photon after/p photon before)

Answer 1

you used this? http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compeq.html#c1

Answer 2

You have to use Compton scattering and momentum
\[\lambda ' - \lambda = \lambda_c(c - \cos \phi) \] and \[p = \frac{ h }{ \lambda }\]

Answer 3

\[\lambda ' = \lambda + \left( \frac{ h }{ mc } \right)(1-\cos \phi)\] then from here you can use conservation of momentum to find the angle

Answer 4

You need it for both x and y components

Answer 5

Yeah, I am looking for the angle of the electron and it's momentum though. Is it correct that the tan inverse of the final potential energy of the photon over its initial energy gives the angle of the electron? \[P'/P=\tan \phi \]

Answer 6

If that's how you've labelled your components, its just like how you did momentum problems back in the day, you use Pythagorus for the momentum of the electron and tan theta for for the direction.

Answer 7

So if you can show me your set up, it would be easier for me to see what exactly you're doing

Answer 8

\[P=(hc/\lambda) (1/C) = 1240eV/.144nm = 8611eV/C \]
\[\lambda' -.144 = (1240/(.511*10^6))(1-\cos60) ........\lambda' = 0.145nm\]
\[P'=1240/0.145=8539eV/c\]
\[P'/P=(Pelectronsin \phi)/(Pelectron \cos \phi)=\tan\]
\[\phi=44.78\]degrees but that isn't right :( Also thank you for the help I have been stuck on this for a long time

Answer 9

Nice try! Alright, so lets go over this problem.

Answer 10

So we have the equations of Compton scattering and of the momentum. So what we have to do here is find the scattered wavelength first, as I showed above, \[\lambda ' = \lambda + \left( \frac{ h }{ mc } \right) (1-\cos \phi)\] so we need this scattered wavelength before we can use conservation of momentum. But, remember as I was saying before we need to break it into components if we want to use conservation of momentum. So we have \[\huge P_{photon}+ P_{electron} = P'_{photon}+P'_{electron}\] momentum conservation. So we'll have it for the x direction and the y direction, can you try that out?

Answer 11

\[\lambda'=\lambda+hc/mc^2 (1-\cos60) \] I multiplied top and bottom by c so its easier to compute. then plugging it in: \[\lambda' = 0.144nm + (1240eV/(0.511*10^6 eV)) (1-\cos60)=0.145nm\]
Since the electron is at rest to start, \[Pphoton=P'photon +Pe'\]
That was teh first step I did

Answer 12

Well, you're thinking about it wrong, it's not like conservation of energy. If you remember using Newton's laws, remember how we always break it into components for the force, we treat momentum the same way.
So, lets look at the x direction, maybe you should draw a picture it will help you understand.
\[\huge x: \frac{ hf }{ c } + 0 = \frac{ hf' }{ c }\cos(\phi)+p_{ex}\]
\[\huge y: 0 + 0 = - \frac{ hf' }{ c } \sin (\theta) + p_{ey}\]

Answer 13

Now you can solve for Pex and Pey and then use Pythagorus and tan ratio to find the angle.

Answer 14

Does that make sense?

Answer 15

well using that I get \[4.6*10^-24 = 4.57*10^-24 \cos \phi+P'e\]
\[P'ey=3.96*10^-24\]

Answer 16

How do I move forward from there?

Answer 17

I'm not checking your numbers, but it does look pretty good, so solve for Pex, and then we just have to use Pythagorean theorem, to find the resultant vector Pe.

Answer 18

\[P_e = \sqrt{(P_{ex})^2+(P_{ey})^2}\]

Answer 19

\[\sqrt{(4.6*10^-24 -4.57*10^-24 \cos \phi)^2+(3.96*10^-24)^2}\]
What about the cos though?

Answer 20

Which cos?

Answer 21

Oh that's cos60

Answer 22

cos60 and sin60 is what you're using since we just broke it into components

Answer 23

For both I use an angle of 60? The electron should have a different angle than the photon?

Answer 24

No, we're looking for the angle of the linear momentum of the electron just after the collision with the photon, that's the point of the question

Answer 25

So we have to solve that still

Answer 26

That's why we use \[\tan \theta = \frac{ P_{ey} }{ P_{ex} }\]

Answer 27

Ah ok, well plugging all that in I get 4.59*10^-24

Answer 28

Know the units? :)

Answer 29

should be kgm/s

Answer 30

Good!

Answer 31

Ok, so thats our momentum for the electron just after the collision with the photon, now we just need the angle, as I showed above, solve for theta.

Answer 32

=59.7 degrees! :D and its right! wow thanks soooo much for taking the time to help, I have been stuck on that problem for a long time!

Answer 33

There we go :)

Answer 34

Np ^.^, that's why I'm here right hehe