Photo-electric effect, Compton Scattering

[mm2424]

Homework Statement

What is the maximum kinetic energy of electrons knocked out of a thin copper foil by Compston scattering of an incident beam of 17.5 KeV rays? Assume the work function is negligible.

Homework Equations

Δλ = h/mc (1-cosθ)

The Attempt at a Solution

I reasoned that the greatest energy transfer to an electron will occur when the x-ray rebounds at 180 degrees, in which case the change in wavelength is 4.85 x 10^-12 m.

I figured that the wavelength of the x-rays increases by this amount, thereby decreasing in energy. I thought I could therefore take this change in wavelength and calculate the energy associated with it using E = hc/Δλ, and I got E = 256 KeV.

However, the answer key requires that you calculate the wavelength of the original x-ray, add Δλ, then calculate the energy and deduct the original energy. It yields a different answer, 1.1 KeV.

I don't understand why you can't say that the energy loss associated with the increase in the wavelength of the x-ray is completely transferred to the electron and then be done with it. What am I missing? Thanks!

 

[mm2424]

Does anybody have any insight into what I'm doing wrong here? I'd appreciate any and all help :).

 

[Redbelly98]

The Attempt at a Solution

I reasoned that the greatest energy transfer to an electron will occur when the x-ray rebounds at 180 degrees, in which case the change in wavelength is 4.85 x 10^-12 m.

I figured that the wavelength of the x-rays increases by this amount, thereby decreasing in energy.

So far, so good.

I thought I could therefore take this change in wavelength and calculate the energy associated with it using E = hc/Δλ, and I got E = 256 KeV.

Not quite. First, note that it's impossible for a 17.5 keV photon to lose 256 keV of energy.

You have two energies, E₁ = hc/λ₁ and E₂ = hc/λ₂.

The energy difference E₁-E₂ is the difference between the hc/λ terms, not hc/Δλ. What you did is equivalent to saying

(1/5) - (1/3) = 1/(5-3) = 1/2,​

which is not true.

Can you take it from here?

However, the answer key requires that you calculate the wavelength of the original x-ray, add Δλ, then calculate the energy and deduct the original energy. It yields a different answer, 1.1 KeV.

Yes, that is the basic idea.

 

[mm2424]

Thanks, that makes sense!

 

[asteriatic]

I understand your reasoning for calculating the maximum kinetic energy of the electrons using the change in wavelength. However, in the context of the photoelectric effect and Compton scattering, it is important to consider the energy conservation principle.

In the photoelectric effect, the energy of the incident photon is completely transferred to the electron, resulting in the electron being ejected with a kinetic energy equal to the energy of the photon minus the work function of the metal. In Compton scattering, the energy of the incident photon is partially transferred to the electron, resulting in a decrease in energy of the scattered photon.

In this case, the incident beam of 17.5 KeV rays has an energy of 17.5 KeV, which is completely transferred to the electron in the photoelectric effect. However, in Compton scattering, the scattered photon will have a lower energy due to the energy transfer to the electron. Therefore, the maximum kinetic energy of the electrons will not be 17.5 KeV, but rather a lower value.

To calculate the maximum kinetic energy of the electrons, we must first calculate the energy of the scattered photon. This can be done by subtracting the change in energy (which is related to the change in wavelength) from the energy of the incident photon. Once we have the energy of the scattered photon, we can use the photoelectric effect equation to calculate the maximum kinetic energy of the electrons.

In summary, while your reasoning for calculating the maximum kinetic energy using the change in wavelength is sound, it does not take into account the energy conservation principle and the fact that the incident photon in Compton scattering will have a lower energy due to the energy transfer to the electron.