91Ó°ÊÓ

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

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 $x$the initial direction of the photon 's momentum. The total momentum of the system before the collision is equal to the momentum of the $\mathrm{γ}$photon:

$p=\frac{E_{\mathrm{γ}}}{c},$

Here, $E_{\mathrm{γ}}=350\text{keV}$The total initial energy is equal to the energy of the photon plus the rest energy of the electron

$E=E_{\mathrm{γ}}+m_e{c}^2\text{.}$

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:

$p=p_e^{\text{'}}−p_{{\mathrm{γ}}^{\text{'}}}^{\text{'}}$

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

$E\text{'}=E{\text{'}}_{\mathrm{γ}}+E{\text{'}}_e$

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

${\mathrm{λ}}^{\text{'}}=\mathrm{λ}+\frac{h}{m_{e}c}\left(1−\cos \mathrm{π}\right)=\mathrm{λ}+\frac{2h}{m_{e}c},$

where we have used the fact that$\mathrm{θ}=\mathrm{π}$for backscattering. So

$E_{\mathrm{γ}}^{\text{'}}=\frac{hc}{{\mathrm{λ}}^{\text{'}}}=\frac{E_{\mathrm{γ}}{m}_e{c}^2}{m_{\mathrm{Ï\mu }}{c}^2+2E_{\mathrm{γ}}}$

The two expressions for $E$derived:

$E_{\mathrm{γ}}+m_e{c}^2=E_{\mathrm{γ}}^{\text{'}}+E_e^{\text{'}}=\frac{E_{\mathrm{γ}}{m}_e{c}^2}{m_e{c}^2+2E_{\mathrm{γ}}}+E_e$

Solving for the energy of the electron we finding

$E_e=m_e{c}^2+\frac{2E_{\mathrm{γ}}^2}{2E_{\mathrm{γ}}+m_e{c}^2}$

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

$E_e=\frac{m_e{c}^2}{\sqrt{1−\frac{v^2}{c^2}}}$

Equating these two expressions as:

$m_e{c}^2+\frac{2E_{\mathrm{γ}}^2}{2E_{\mathrm{γ}}+m_e{c}^2}=\frac{m_e{c}^2}{\sqrt{1−\frac{v^2}{c^2}}}$

This yields

$\sqrt{1−\frac{v^2}{c^2}}=\frac{m_e{c}^2\left(2E_{\mathrm{γ}}+m_e{c}^2\right)}{m_e{c}^2\left(2E_{\mathrm{γ}}+m_e{c}^2\right)+2E_{\mathrm{γ}}^2}$

Squaring this and rearranging terms as:

$\frac{v^2}{c^2}=1−{\left(\frac{m_e{c}^2\left(2E_{\mathrm{γ}}+m_e{c}^2\right)}{m_e{c}^2\left(2E_{\mathrm{γ}}+m_e{c}^2\right)+2E_{\mathrm{γ}}^2}\right)}^2$

role="math" localid="1651139373123" $v=c\frac{2E_{\mathrm{γ}}\left(E_{\mathrm{γ}}+m_e{c}^2\right)}{2{E^2}_{\mathrm{γ}}+2m_e{c}^2{E}_{\mathrm{γ}}+{m^2}_e{c}^4}\text{.}$

Plugging $E_{\mathrm{γ}}=350keVand{m}_e{c}^2=511keV$as:

$=0.70c=2.1×{10}^{8}m/s$