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Compton scattering is relevant not only to x-ray photons but, even more so, to higher energy gamma-ray photons. Suppose a350keV gamma-ray photon backscatter (i.e., is scattered back toward the source) from a free electron. Afterward, what is the electron’s velocity in m/s?

Short Answer

Expert verified

The electron’s velocity in m/sis2.1×108m/s.

Step by step solution

01

Given Information

We need to find the electron’s velocity in m/s.

02

Simplify

We will use the energy and the momentum conservation to solve this problem and also the relativistic relations for energy and momentum. Direct the axis xthe initial direction of the photon 's momentum. The total momentum of the system before the collision is equal to the momentum of the γphoton:

p=Eγc,

Here, Eγ=350keVThe total initial energy is equal to the energy of the photon plus the rest energy of the electron

E=Eγ+mec2.

After the collision, the total momentum is equal to the momentum of the electron minus the momentum of the photon since the vector its momentum points in the negative direction:

p=pe'pγ''

The total energy after the collision is equal to the energy of the scattered photon plus the energy of the scattered electron

E'=E'γ+E'e

From the formula of the Compton scattering we have that the wavelength of the scattered photon is:

λ'=λ+hmec1cosπ=λ+2hmec,

where we have used the fact thatθ=πfor backscattering. So

Eγ'=hcλ'=Eγmec2mϵc2+2Eγ

03

Calculation

The two expressions for Ederived:

Eγ+mec2=Eγ'+Ee'=Eγmec2mec2+2Eγ+Ee

Solving for the energy of the electron we finding

Ee=mec2+2Eγ22Eγ+mec2

Also, the total energy of the electron in terms of its speed is

Ee=mec21v2c2

Equating these two expressions as:

mec2+2Eγ22Eγ+mec2=mec21v2c2

This yields

1v2c2=mec22Eγ+mec2mec22Eγ+mec2+2Eγ2

Squaring this and rearranging terms as:

v2c2=1mec22Eγ+mec2mec22Eγ+mec2+2Eγ22

role="math" localid="1651139373123" v=c2EγEγ+mec22E2γ+2mec2Eγ+m2ec4.

Plugging Eγ=350keVandmec2=511keVas:

=0.70c=2.1×108m/s

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