How Is the Maximum Kinetic Energy of Compton Scattered Electrons Calculated?

[niehls]

there is this problem which I'm having problems solving.

An X-ray beam has an energy of 40keV. Find the maximum possible kinetic energy of Compton scattered electrons.

The electron is initially at rest.
I go at it this way.
For maximum momentum to be delivered from the photon to the electron, the collision must be straight on, reflecting the photon by an angle 180 degrees.
Photon momentum:

before collision: p = E/c
after collision: p = -E/c
if positive direction is along the photon's initial path.
This means the difference in momentum is 2E/c. This momentum must be transferred to the electron (conservation of momentum). Using K = E_total - mc^2, p = 2E/c and
E_total^2 = (pc)^2 + (mc^2)^2

This yields K = 6.22keV. The correct answer is 5.47 keV. Could someone please help and point me in the right direction...
thanks

 

[dextercioby]

HINT:Treat it like a regular relativistic collision/scattering problem.Or maybe that's already done in your textbook.If so,they must have given the formula for KE of the scattered electron.Maximize it wrt to parameters (the scattering angles) and find the maximum value.If u don't have the formula in your textbook,calculate it applying the fundamental law of energy-momentum conservation.That should not be too difficult.

Daniel.

 

[niehls]

i've been working on this but i seem to be stuck. i can't find any function for the kinetic energy of the electron to maximize. I've applied the energy/momentum conservation already deriving the equations above but as i said, i seem to be stuck. could you please elaborate? even more helpful; what's wrong with my above reasoning?

 

[marlon]

Hi, you need to use the formula for the compton scattering :

$$\Delta\lambda = \frac{h}{mc} (1-cos(\theta))$$

the shift is maximal when theta is equal to 180° so that $$\Delta\lambda = 4.8 * 10^{-12}pm$$

E is energy of incident photon and E' is energy of scattered photon.

$$\Delta\lambda = {\lambda}' - \lambda$$
Now the fraction of ebergy that is transferred to the electrons is equal to :
$$\frac{E-E'}{E} = \frac{{\lambda}' - \lambda}{{\lambda}'}$$

or this is equal to : $$\frac{\Delta\lambda}{\lambda + \Delta\lambda}$$

This fraction is about 13 % of the 40 keV...you can calculate this and your problem is solved...

regards
marlon

 

[Janitor]

before collision: p = E/c
after collision: p = -E/c

I will merely add that in the lab frame, where the electron is initially at rest, I would expect the magnitude of the reflected photon's momentum to be less than the magnitude of the initial photon momentum. I think that your answer came out too large because of this invalid assumption. (In the CMS frame it would be true that final photon momentum magnitude = initial photon momentum magnitude, but of course in the CMS frame the initial photon energy would be smaller than the energy given in the problem statement.)

 

[niehls]

E is energy of incident photon and E' is energy of scattered photon.

$$\Delta\lambda = {\lambda}' - \lambda$$
Now the fraction of ebergy that is transferred to the electrons is equal to :
$$\frac{E-E'}{E} = \frac{{\lambda}' - \lambda}{{\lambda}'}$$

or this is equal to : $$\frac{\Delta\lambda}{\lambda + \Delta\lambda}$$

This fraction is about 13 % of the 40 keV...you can calculate this and your problem is solved...

I was just wondering about equating the relative energy lost by the photon to the relative wavelength shift.
$$\frac{E-E'}{E} = \frac{{\lambda}' - \lambda}{{\lambda}'}$$
Haven't had coffee yet so i hope you'll excuse me. Please tell me why this is an equality.

 

[Nylex]

the shift is maximal when theta is equal to 180° so that $$\Delta\lambda = 4.8 * 10^{-12}pm$$

You made a mistake here, surely? You mean 4.8 x 10^-12 m or 4.8 pm.

 

[marlon]

Nylex, thanks for the correction...

it is indeed 4.8pm

regards
marlon

 

[niehls]

marlon, could you please explain your above equality?

cheers

 

[marlon]

Niehls,...

you know that E = hv (v frequence and h Planck-constant) This is the first deBroglie-relation connection energy to frequence (particle-wave-duality.)

Also $$\lambda * v = c$$and v is again frequence for EM-radiation (photons)

Now in (E-E')/E replace E by hv : (hv-hv')/hv...h gets out and then replace v by $$c/ {\lambda}$$

This yields :

$$\frac{c/{\lambda} - c/{\lambda}'}{c/{\lambda}}$$

Then make the denominators equal and all is done...

regards
marlon

 

[Janitor]

I just now worked the problem, using the relativistic momentum for the photon (obviously!) but using the nonrelativistic momentum for the electron.

E_i= initial photon energy in lab frame = 40 keV
E_f=final photon energy in lab frame
v=final electron speed in lab frame

Momentum conservation and the fact that for massless particles momentum = E/c gives:

E_i/c = mv - E_f/c

Energy conservation gives:

E_i = E_f + (1/2)mv^2.

The latter can be solved for v to give:

v= sqrt[(2/m)(E_i-E_f)].

Substituting this result for v into the first equation, and writing it as a quadratic equation in the unknown E_f:

E_f^2 + (2E_i + 2mc^2)E_f + (E_i^2 - 2mc^2E_i) = 0.

Then use

E_i = 40 keV
mc^2 = 511 keV

Discarding the negative solution leaves:

E_f = 34.56 keV.

So the electron energy is 40 keV - 34.56 keV = 5.44 keV in this semi-relativistic calculation. That is within a half percent of the answer that niehls says is correct.