Which scattering angle corresponds to the largest Compton shift and why?

[SpY]]

Homework Statement

An electron initially at rest is scattered by a photon.
a) Which scattering angle corresponds to the largest Compton shift and why?
b) At what minimum photon energy can half of the photon energy be transferred onto the electron?

Homework Equations

\Delta \lambda = \lambda_2 - \lambda_1 = \frac {h}{m_e c} (1 -cos\theta)
where \Delta \lambda is the compton shift, \lambda_2 and \lambda_1 are the final and initial photon wavelengths, m is the mass of the electron, c the speed of light, and \theta the scattering angle.

The Attempt at a Solution

a) Pretty straightforward, for a maximum shift \delta \lamba the (1 -cos\theta) must be a maximum, ie. cos\theta) = 0 which happens at 90 degrees.

b) I might be confused in understanding the question. Way I figured, is that the the initial photon kinetic energy is shared equally between the reflected photon and electron. So conservation of mechanical energy (without potential energy, electron initially at rest, K = hf for photon and K= 0.4mv² for electron)
h \frac {c}{\lambda_1} = h \frac {c}{\lambda_2} + 0.5 m v^2
So since the final energy of the photon is half the initial, that means the reflected photon and electron have equal resultant kinetic energies.
h \frac {c}{\lambda_1} = 2 h \frac {c}{\lambda_2} = m v^2

I'm still not getting what "minimum" is

 

[da_nang]

You might want to look a little further into a). You find the maximum of a function by finding the roots of the derivative.

As for the second question, and I'm not quite sure, but I believe you're supposed to write the initial energy of the photon as a function of θ, then find the minimum energy you can get.

 

[vela]

Homework Statement

An electron initially at rest is scattered by a photon.
a) Which scattering angle corresponds to the largest Compton shift and why?
b) At what minimum photon energy can half of the photon energy be transferred onto the electron?

Homework Equations

\Delta \lambda = \lambda_2 - \lambda_1 = \frac {h}{m_e c} (1 -cos\theta)
where \Delta \lambda is the compton shift, \lambda_2 and \lambda_1 are the final and initial photon wavelengths, m is the mass of the electron, c the speed of light, and \theta the scattering angle.

The Attempt at a Solution

a) Pretty straightforward, for a maximum shift \delta \lamba the (1 -cos\theta) must be a maximum, ie. cos\theta) = 0 which happens at 90 degrees.

Pretty straightforward, but you got it wrong. (1-cos θ) is equal to 2sin² θ/2. When does the latter achieve its maximum?

b) I might be confused in understanding the question. Way I figured, is that the the initial photon kinetic energy is shared equally between the reflected photon and electron. So conservation of mechanical energy (without potential energy, electron initially at rest, K = hf for photon and K= 0.4mv² for electron)
h \frac {c}{\lambda_1} = h \frac {c}{\lambda_2} + 0.5 m v^2
So since the final energy of the photon is half the initial, that means the reflected photon and electron have equal resultant kinetic energies.
h \frac {c}{\lambda_1} = 2 h \frac {c}{\lambda_2} = m v^2

I'm still not getting what "minimum" is

The righthand side of the formula for the change in wavelength says it's limited to a certain range that depends on the mass of the electron; it will be on the order of hc/mc², which is completely independent of the initial energy of the photon. So it's possible that only for a certain range of photon energies will the change in wavelength be consistent with the photon losing half its initial energy. The problem is asking you to find the lower limit of this range.

You might have noticed I threw in an extra factor of c into both the numerator and denominator. It's often useful to do this because certain combinations of constants appear frequently, and you can save a lot of time if you memorize them. In this case, you have hc, which is approximately 1240 MeV-fm (this combo is useful when you want to convert between energy and wavelength for photons), and mc², which is 0.511 MeV, the rest energy of the electron. So hc/mc² = 2430 fm = 2.43x10^-12 m.

 

[SpY]]

sigh silly me... I was looking at when the function (1-cosθ) and not it's derivative (sinθ). Which is obviously zero at 180 degrees! That then makes more sense in how I got to the second equation, which is the compton shift for 180 degrees (ie. for half the energy to be transferred in the collision, it must be at 180 degrees) so

\lambda_2 - \lambda_1 = \frac {2h}{m_e c} = 4860 fm (θ=180)

Combining with \frac {hc}{\lambda_1} = 2 h \frac {c}{\lambda_2} I get E_\lambda_1 = \frac {2hc}{\lambda_1 + 4860fm}
which doesn't help much.

The question as I see it, "At least how much energy must the photon have to lose half" seems redundant, and I don't see how the mass of the electron can be used because it's rebound velocity is another unknown. Am I correct that for whatever energy the photon has, if it collides at 180 degrees, half of it's initial energy will be transferred to the electron initially at rest? (or is my assumption of 180degrees the problem)

 

[vela]

No, the photon scattering at 180 degrees does not necessarily mean that it has lost half its energy. In fact, that would only be true for one energy. At some other energies, it's just not possible, and for other energies, losing half would occur at different angles.

 

[SpY]]

For a scattered photon to have half the energy of the initial photon, it must have double the wavelength, right? (2\lambda_2 = \lambda_1)
(I'm assuming θ=180 because that seems to give the minimum energy to me)

\lambda_2 - \lambda_1 = \frac {2h}{mc}
\lambda_1 = \frac {2h}{mc}

E_\lambda_1 = \frac {hc}{\lambda_1} = \frac {mc hc}{2h} = 0.5mc^2

For which all values are known, I get the energy 0.256MeV?

 

[vela]

For a scattered photon to have half the energy of the initial photon, it must have double the wavelength, right? (2\lambda_2 = \lambda_1)
(I'm assuming θ=180 because that seems to give the minimum energy to me)

There's no need to assume. That's what part (a) of the problem told you.

\lambda_2 - \lambda_1 = \frac {2h}{mc}
\lambda_1 = \frac {2h}{mc}

E_\lambda_1 = \frac {hc}{\lambda_1} = \frac {mc hc}{2h} = 0.5mc^2

For which all values are known, I get the energy 0.256MeV?

That's correct.