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

$$\Delta \lambda \neq \frac{hc}{\Delta E}$$

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

 

[baba89]

Your understanding is correct. In a Compton collision, the angle of deflection is determined by the equation (delta)lambda=(h/mec)(1-cos(theta)), where (delta)lambda is the change in wavelength, h is Planck's constant, me is the mass of the electron, and c is the speed of light. This equation can be rearranged to solve for the angle theta, which in this case is 180 degrees.

Using this value of theta, we can calculate the final energy of the photon using the equation E=hc/(delta)lambda. The initial energy of the photon is given as 30 keV, so we can plug in the values and solve for the final energy, which would be approximately 29.97 keV.

As you mentioned, due to conservation of energy, the energy lost by the photon must equal the energy gained by the electron. Therefore, the final energy of the electron would also be approximately 29.97 keV. This is because the initial energy of the electron was at rest, so all of the energy gained from the photon is transferred to the electron.

Overall, your understanding and approach to this problem is correct. Keep up the good work!