Fraction of lost energy in compton scattering

[Magnetic Boy]

Homework Statement

After undergoing through 90° compton scattering, the fraction of energy lost by photon is
a) 10%
b) 20%
c) 50%
d) zero
e) none of these

Homework Equations

∆λ= h/m_oc (1-cosΦ)

The Attempt at a Solution

What i m doing is that, i get scattered photon energy and subtracting it from the total and dividing by the total. But it seems unsolvable as wave length of the photon before scattering not given

 

[malawi_glenn]

what is the relation between the Energy of a photon and it's wavelength?

 

[malawi_glenn]

If the energy of the photon before the scattering is denoted E and the energy of the photon is denoted E' what is the expression for "the fraction of energy lost"

 

[Magnetic Boy]

Is there any such relation?? I don't know...

 

[malawi_glenn]

google it

 

[malawi_glenn]

but to answer the question. Look at the numbers given, they are really "nice". No way those can be reproduced by the scattering formula.

 

[Magnetic Boy]

Got the eqn. But it need to have scattered energy frequency. Which is not given in the problem. Does this mean the option "none of above is correct"?

 

[James R]

For a photon (light), frequency and wavelength are related.

 

[Magnetic Boy]

For a photon (light), frequency and wavelength are related.

Yes. But neither of them is given

 

[Delta2]

Frequency and Energy of photon also are related E=hf where h plank's constant.

 

[Magnetic Boy]

QUOTE="Delta², post: 5517626, member: 189563"]Frequency and Energy of photon also are related E=hf where h plank's constant.[/QUOTE]
I know it very well. But look at the question. Only angle of scattering is given. Does not it mean that we cannot find fraction of lost energy? I just want to confirm.. (or is there some way to find the fractional lost energy)

 

[Delta2]

Frequency and Energy of photon also are related E=hf where h plank's constant.

I know it very well. But look at the question. Only angle of scattering is given. Does not it mean that we cannot find fraction of lost energy? I just want to confirm.. (or is there some way to find the fractional lost energy)

Seems to me you are right, we have to know the wavelength (or frequency) of the photon before scattering.

 

[dpopchev]

The Attempt at a Solution

What i m doing is that, i get scattered photon energy and subtracting it from the total and dividing by the total. But it seems unsolvable as wave length of the photon before scattering not given

Should not have the need for initial energy: https://www.hep.wisc.edu/~prepost/407/compton/compton.pdf

EDIT: made a calculation mistake, have no idea how to approach this questions

 

[James R]

Got the eqn. But it need to have scattered energy frequency. Which is not given in the problem.

Just to check, what equation did you get?

 

[Magnetic Boy]

Just to check, what equation did you get?

I got ΔE/E= (E'/mc²)(1-cosΦ)

 

[dpopchev]

I found this pdf where question 1 is similar: http://users.wpi.edu/~physics/ph1130c06/Images/ppset05.PDF

 

[James R]

I got ΔE/E= (E'/mc²)(1-cosΦ)

That looks right.

I agree that you'd need to know the wavelength or frequency of the incoming photon in order to get a numerical answer.

 

[Magnetic Boy]

That looks right.

I agree that you'd need to know the wavelength or frequency of the incoming photon in order to get a numerical answer.

Thanks. So the answer is "none of these". Now i am sure. Some one answered it 50%. And i were really confused about that.

 

[James R]

Thanks. So the answer is "none of these". Now i am sure. Some one answered it 50%. And i were really confused about that.

Well, I supposed it could be 50%, or 10% or 20%, if the incoming photon energy was whatever is necessary to get those values. We know it can't be zero, because incoming photon must lose energy if it is scattered at any angle other than 0 degrees. Your formula has $\Delta E/E = E'/mc^2$, where $\phi=90^\circ$, and $E'$ can't be zero.

Really, there should be an "(a), (b) or (c)" option.