Question

hello, how can we obtained this equation from compton scattering equation
cot =(1+&) tan 1/2

Answer 1

Hi.
Equation needs to be written down "explicitly". I recognise the words compton etc, but the equation is a bit low on "content", as far as I can see.
the more info we get, the more chance of helping solve the problem ...

Answer 2

\[\cot \theta =(1+\alpha) \tan \phi/2\]

Answer 3

@osprey

it's this, isn't it ?!?!

\(\cotφ = \left(1+ \dfrac{hf}{m_ec^2} \right) \tan \left(\dfrac{θ}{2} \right)\)

i'd say to the OP: drawing it and stuff is a good idea....

and i reckon the solution can be ripped out from this:
http://physics.tutorvista.com/modern-physics/compton-scattering.html

IOW: a load of shiττy algebra.

the original paper, so Google says ....
http://journals.aps.org/pr/pdf/10.1103/PhysRev.21.483

the manual \(\nu\)'s on the type-set are soo cute :-))

Answer 4

@IrishBoy123 it looks like a bit of the usual "symbols car crash" here. And there's the tiny point of what the Compton effect actually is, and its significance, and Compton's Nobel prize for it.
I think that it demonstrated that an incoming EM wave could interact with a "host" particle in the same way that an incoming particle would do so in terms of collisions. Add a bit of quantum, and a bit of rel and both Q and rel "gain credence" ...
I "think". But I'm awash with symbols at the moment ...

Answer 5

Continuing with the Wiki derivation started here:

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

.....and simply resolving momentum along, and orthogonal to, the path of the incoming photon \(\gamma\) .....

ALONG

\(\dfrac{hf}{c} = \dfrac{hf'}{c} \cos \theta + mv \cos \phi \qquad \star\)

PERPENDICULAR

\(0 = \dfrac{hf'}{c} \sin \theta - mv \sin \phi \qquad \triangle\)

so

\(\cot \phi = \dfrac{ (\dfrac{hf}{c} - \dfrac{hf'}{c} \cos \theta)/mv } { -(\dfrac{hf'}{c} \sin \theta )/(-mv)} \)

\(\implies \cot \phi = \dfrac{ \dfrac{f}{f'} - \cos \theta } { \sin \theta } = \dfrac{ \dfrac{\lambda'}{\lambda} - \cos \theta } { \sin \theta } \)

From the main result:

\( \lambda '-\lambda ={\dfrac {h}{m c}}(1-\cos \theta )\)

\( \implies \dfrac{\lambda'}{\lambda} = 1 + {\dfrac {h}{ \lambda m c}}(1-\cos {\theta })\)

\(= 1 + {\dfrac {hf}{ m c^2}}(1-\cos {\theta })\)



\(\implies \cot \phi = \dfrac{ 1 + {\dfrac {hf}{ m c^2}}(1-\cos {\theta }) - \cos \theta } { \sin \theta } \)

\(= \dfrac{ \left(1 + {\dfrac {hf}{ m c^2}} \right)(1-\cos {\theta }) } { \sin \theta } \)

\(= \left(1 + {\dfrac {hf}{ m c^2}} \right) \tan \dfrac{\theta}{2}\)

...using the tan half angle identity