Compton scattering photons per radian vs steradian at 150° and 90°

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Homework Statement



In an experiment there was found equal compton scattered photons per steradian in the direction 150* as in 90*

Homework Equations



What relation in number of scattered photons in 150* and 90* would you get if you define number of photons per radians instead?

The Attempt at a Solution



I feel that this one is easy, but I don't know how to start with this.

Thanks!
 
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hi abotiz! :smile:

(have a degree: ° :wink:)

be logical …

start with the definition …

what is the definition of steradian? :wink:
 
[tex]\stackrel{d\sigma}{d\theta}[/tex]=[tex]\stackrel{d\sigma}{d\Omega}[/tex]*[tex]\stackrel{d\Omega}{d\theta}[/tex]

where [tex]\stackrel{d\Omega}{d\theta}[/tex] = 2[tex]\pi[/tex]sin[tex]\theta[/tex]

so the answer is sin90/sin150 i.e = 2

Right?

Thanks
 
Last edited:
hi abotiz! :smile:

(just got up :zzz: …)
abotiz said:
[tex]\stackrel{d\sigma}{d\theta}[/tex]=[tex]\stackrel{d\sigma}{d\Omega}[/tex]*[tex]\stackrel{d\Omega}{d\theta}[/tex]

where [tex]\stackrel{d\Omega}{d\theta}[/tex] = 2[tex]\pi[/tex]sin[tex]\theta[/tex]

so the answer is sin90/sin150 i.e = 2

Right?

Thanks

sorry, i don't understand that at all :confused:

what are you using as the definition of steradian?
 
Hi!

I found that definition in the Literature we are using.

Seems to be derived from [tex]\sigma[/tex] = [tex]\int\stackrel{d\sigma}{d\Omega}[/tex]*[tex]\Omega[/tex] ( total scattering cross section )

Anyways, I think my answer is right, perhaps the definitions we are using is based for a special case or something, which I don't know anything about :-p

Thanks
 
That's the formula for the integral of the differential cross-section (σ is ordinary area in m2, Ω is angular area in steradians) …

where do radians come into it? :confused:
 
Beats me.

But its not that weird, if you would take the derivative of that equation and divide both side with the differential angle, or what you call it ( dtheta) you would get the equation I used. Anyways, Thanks