Kinetic energy of a recoiled electron when backscattered

[StephenD420]

Show that for very high energy photons, the kinetic energy of the recoiling electron approaches
KE= Ephoton - 255.5 KeV for back-scattering, phi = 180 degrees.

I know that

λ'/hc = λ/hc + (1-cos phi)/m0c^2
and since E = hc/λ
E' = E + m0c^2/(1-cos phi)
when phi is 180 degrees
E' = E + m0c^2/2
and the resting energy of the electron is m0c^2 = .511 MeV
so
E' = E + 255.5KeV
so E = Ephoton - 255.5Kev

Is this right?? I am not sure about the last step where Ephoton = E'?

Thanks.
Stephen

 

[vela]

Show that for very high energy photons, the kinetic energy of the recoiling electron approaches
KE= Ephoton - 255.5 KeV for back-scattering, phi = 180 degrees.

I know that

λ'/hc = λ/hc + (1-cos phi)/m0c^2
and since E = hc/λ
E' = E + m0c^2/(1-cos phi)
when phi is 180 degrees
E' = E + m0c^2/2
and the resting energy of the electron is m0c^2 = .511 MeV
so
E' = E + 255.5KeV
so E = Ephoton - 255.5Kev

Is this right?? I am not sure about the last step where Ephoton = E'?

Thanks.
Stephen

No, it's not correct because 1/a = 1/b + 1/c doesn't imply a = b + c, which is what you did when you went from wavelength to energy.

 

[StephenD420]

ok so...what do I do?

 

[StephenD420]

Ok
1/E'=1/E +(1-cos phi)/.511Mev
now if you inverse both sides what do you get then

 

[StephenD420]

ok so I put this into Mathematica and got
In[2]:= simplify ((1/a)^-1 == (1/b)^-1 + (1/c)^-1)



Out[2]= simplify (a == b + c)

so
E' = E/(1+(E/m0c^2)*(1- cos phi))

Which is my answer when phi = 180 degrees and m0c^2 = .511MeV

so what do you think?

 

[vela]

You seem to think E' is the kinetic energy of the electron. It isn't.

 

[StephenD420]

ok so can you give me a hint then...

 

[vela]

ok so I put this into Mathematica and got
In[2]:= simplify ((1/a)^-1 == (1/b)^-1 + (1/c)^-1)



Out[2]= simplify (a == b + c)

If you have $\frac{1}{a} = \frac{1}{b}+\frac{1}{c}$, it follows that $a = \left[\frac{1}{b}+\frac{1}{c}\right]^{-1}$. You had already repeated the mistake I pointed out when you entered the above into Mathematica.

First thing you want to do is get things straight in your head. What does E represent? What does E' represent? What quantity are you trying to solve for? How can that be expressed in terms of E, E', and other variables in the problem?

 

[StephenD420]

I am just trying to show that with high energy photon the recoiling KE approaches KE = Ephoton - 255.5 KeV

E' is Ephoton
and E is E photon before collision

now since I get the right answer...I do not know how else to get the right answer...any help would be appreciated and (Δλ/hc)^(-1) = hc/Δλ = ΔE = E' - E = (1/m0c^2 *(1-cos phi))^(-1) = m0c^2/(1-cos phi).

and if phi = 180 degrees and m0c^2 = .511 Mev
ΔE = .511MeV/2 = .2555 KeV
so I really do not know what you are getting at...

 

[vela]

What I'm getting at is that you're not doing the basic algebra correctly. You have
$$\frac{hc}{\Delta \lambda} = \frac{hc}{\lambda' - \lambda}$$ and
$$ E'-E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}.$$ You're claiming those two are equal because
$$\frac{hc}{\lambda' - \lambda} = \frac{hc}{\lambda'} - \frac{hc}{\lambda}.$$ Note this is akin to saying $\frac{1}{3-1}$ is equal to $\frac{1}{3}-\frac{1}{1}$. It's obviously not correct.

 

[StephenD420]

and I am saying ok since I getting the same numbers the professor wants, then if that is not right please point me in the right direction as I do not see how the professor got the numbers otherwise if not the rest energy of the electron over (1-cos phi) = 2