A 10.0-keV X-ray photon is successively scattered by two...

A $10.0-\mathrm{keV}$ X-ray photon is successively scattered by two free electrons initially at rest. In the first case, it is scattered through an angle of $41^{\circ}$; the second scattering is through an angle of $72^{\circ} .$ (a) What is the final photon energy? (b) How much kinetic energy does each electron receive?

A $10.0-\mathrm{keV}$ X-ray photon is successively scattered by two free electrons initially at rest. In the first case, it is scattered through an angle of $41^{\circ}$; the second scattering is through an angle of $72^{\circ} .$ (a) What is the final photon energy?
(b) How much kinetic energy does each electron receive?

Key Concepts

Compton Scattering

This is the physical phenomenon where a photon collides with a free or loosely bound electron, resulting in a change in the photon's direction and energy. It is central to the problem, as it explains how an X-ray photon interacts with electrons and loses energy by transferring some of it to the electrons.

Compton Wavelength Shift

This concept is embodied in the Compton formula, ?? = (h/m_ec)(1 - cos ?), which quantifies the change in the wavelength of the photon as a function of the scattering angle. It is crucial for determining how the photon’s energy is reduced after each scattering event by relating the angle of scattering to the increase in the wavelength.

Energy Conservation in Photon-Electron Interactions

This principle states that the total energy before and after the interaction is conserved. In the context of Compton scattering, the initial energy of the photon is divided into the energy of the scattered photon and the kinetic energy imparted to the electron. This concept is essential for calculating the energy distribution after each scattering event.

Momentum Conservation in Photon-Electron Collisions

In addition to energy, momentum is conserved during the scattering process. The momentum conservation laws help determine the directions and magnitudes of the resultant momenta of both the scattered photon and the recoiling electron. This is especially important when analyzing sequential scattering events where the directions of scatter must be taken into account.

Sequential Scattering Events

This involves applying the Compton scattering process more than once, where a photon that has already been scattered by one electron undergoes additional scattering with another. Each scattering event must be analyzed independently using the Compton formula and conservation principles, with the subsequent events taking the modified photon energy and wavelength from the previous interaction as the new starting parameters.

Transcript

00:01 In this question, we have content scattering between x -ray photon with two free electrons.

00:07 So, so incoming photon with a 10 kv energy and then it gets scattered and then there's enough.

00:17 So that's the first scattering, then it hits another electron again and then scatter again.

00:27 So this is e1 and then this is e2.

00:31 2 and lambda 1 2.

00:37 So the first scattering angle the first scattering angle is 41 degrees the second scattering angle is 72 degrees so in this question you want to find a final photon energy and the kinetic energy that each electron receive so to do part a key to find e2 so we use the content scattering equation the content scattering equation okay we say that delta lambda it goes to lambda minus lambda not equals to lambda c times one minus cosine data okay so for the first scattering here we have lambda one minus lambda not sequel to and i c times 1 minus cosine data 1.

01:52 Okay then the second scattering here we have lambda 2 minus lambda 1 goes to lambda c times 1 minus cosine data 2 so we are interested to find lambda 2 because we want to find e 2 so lambda 2 equals to lambda 1 plus lambda c times 1 minus cosine data 2 and lambda 1 is lambda 0 plus lambda c and lambda 1 plus lambda c 1 minus cosine data 1 plus lambda c times 1 minus cosine 2 okay so here we don't have lambda naught so you use hc over e0 okay to represent lambda not and then we have lambda c times 2 minus cosine theta 1 minus cosine 2 2 so now we can substitute a very values.

03:17 So just keep writing and you get there.

03:22 Okay after putting all the numbers and you press calculator and you get the lambda 2 to be 1 .266 times 10 to the negative 10 meters.

03:35 So final photon energy e2 is equal to each c over lambda 2.

03:42 So just put in the relevant numbers...

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