Compton Collision: Photon & Electron 180° Scatter, 30keV Energy

[Quelsita]

Suppose that after a Compton collision between a photon and electron initially at rest, the electron and photon emerge symmetrically (in equal and opposite angles).
If the initial energy of the photon=30 keV, a) what is the angle that corresponds to such scattering? and b) what is the final photon energy? c) final electron energy?

1. If the angle of deflection is symmetric, then theta=180?

2. since we know theta, we can use
(delta)lambda=(h/mec)(1-cos(theta))
where =(h/m_ec) is the compton wavelength=0.02426A
which can be rearranged
hc/(delta)E=(0.02426A)(1-cos(theta))
and plug in theta and the E_iphoton to find the final energy.

3. Due to conservation of energy the total energy lost for the photon must equal that of the energy gained by the electron and we know E_iphoton and E_fphoton this would equal (delta)E_electron.

Is this correct?

 

[alphysicist]

Hi Quelsita,

Suppose that after a Compton collision between a photon and electron initially at rest, the electron and photon emerge symmetrically (in equal and opposite angles).
If the initial energy of the photon=30 keV, a) what is the angle that corresponds to such scattering? and b) what is the final photon energy? c) final electron energy?

1. If the angle of deflection is symmetric, then theta=180?

2. since we know theta, we can use
(delta)lambda=(h/mec)(1-cos(theta))
where =(h/m_ec) is the compton wavelength=0.02426A
which can be rearranged
hc/(delta)E=(0.02426A)(1-cos(theta))

I don't believe this is correct; because

<br /> \Delta \lambda \neq \frac{hc}{\Delta E}<br />

I would suggest finding the final wavelength (not using the \Delta form) and then converting that to energy.