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| | | ### Statement |
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| | − | $14.5.23.$ [Insert the problem statement] |
| | + | $14.5.23.$ A photon of mass $m$ collides with a stationary electron. Determine the mass of the photon and the electron after the collision, in which the photon changes its direction of motion by an angle $\alpha$. |
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| | + |  |
| | + | |
| | | ### Solution |
| | + | This problem deals with the so-called Compton effect: a photon scatters off a free electron, transfers part of its energy to the electron, and its own frequency decreases. This topic is very popular with various olympiad authors. For example, a similar problem appeared in the Belarusian selection for IPhO. |
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| | − | $14.5.23.$ A photon of mass $m$ collides with a stationary electron. Determine the mass of the photon and the electron after the collision, in which the photon changes its direction of motion by an angle $\alpha$. |
| | + | Let us immediately note: the masslessness of the photon and the constancy of the electron's rest mass have not been cancelled. Savchenko uses mass as a measure of the $\textbf{total energy}$ of these particles. Thus, $m = \frac{h\nu}{c^2}$, where $\nu$ is the photon frequency. Or, if using mass in eV, $m = h\nu = E_{photon}$, $m_e = E_{electron}$ — this can significantly simplify the notation. |
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| | + | We will use this notation, although some formulas will look a little unusual. |
| | + | |
| | + | Then the conservation of energy: |
| | + | $$ |
| | + | m_e+m=m'_e+m_{\gamma}\tag{1} |
| | + | $$ |
| | + | Immediately we obtain the "mass" of the electron after the collision: |
| | + | $$ |
| | + | m'_e=m_e+m-m_{\gamma}\tag{2} |
| | + | $$ |
| | + | |
| | + | Relativistic invariant for the electron: |
| | + | $$ |
| | + | m'^{2}_e=p^2_e+m^{2}_e\tag{3} |
| | + | $$ |
| | + | |
| | + | Let the $Ox$ axis be directed along the initial direction of the photon, and $Oy$ perpendicular. Then the angle between the direction of motion of the electron after the collision and $Ox$ is $\varphi$. Conservation of momentum along the axes: |
| | + | $$ |
| | + | Ox: \quad \frac{m}{c}=p_e\cos\varphi+\frac{m_{\gamma}}{c}\cos\alpha\tag{4} |
| | + | $$ |
| | + | $$ |
| | + | Oy: \quad 0=p_e\sin\varphi+\frac{m_{\gamma}}{c}\sin\alpha\tag{5} |
| | + | $$ |
| | + | We are not very interested in where the electron goes (the angle at which the photon is observed is practically important), so we eliminate $\varphi$. To do this, first substitute (3) into (4) and (5), multiply by $c$ and square: |
| | + | $$ |
| | + | (m-m_{\gamma}\cos\alpha)^2=(m'^{2}_e-m^{2}_e)cos^2\varphi\tag{6} |
| | + | $$ |
| | + | $$ |
| | + | m_{\gamma}^2\sin^2\alpha=(m'^{2}_e-m^{2}_e)sin^2\varphi\tag{7} |
| | + | $$ |
| | + | Using the Pythagorean identity: |
| | + | $$ |
| | + | (m-m_{\gamma}\cos\alpha)^2=(m'^{2}_e-m^{2}_e)-m_{\gamma}^2\sin^2\alpha \tag{8} |
| | + | $$ |
| | + | Simplifying, |
| | + | $$ |
| | + | m^2-2mm_{\gamma}\cos\alpha+m_{\gamma}^2=m'^{2}_e-m^{2}_e \tag{9} |
| | + | $$ |
| | + | $(2)\to(9)$ |
| | + | $$ |
| | + | mm_{\gamma}\cos\alpha=m_em_{\gamma}+mm_{\gamma}-mm_e \tag{10} |
| | + | $$ |
| | + | |
| | + | $$ |
| | + | m_{\gamma}=\frac{m}{1+\frac{m}{m_e}(1-\cos\alpha)}<m\tag{11} |
| | + | $$ |
| | + | |
| | + | Note: If we return to the usual energy instead of mass, we obtain the decrease in frequency mentioned at the beginning and get the standard formula for the Compton effect: |
| | + | $$ |
| | + | \frac{1}{\nu'}-\frac{1}{\nu}=\frac{h}{m_e c^2}(1-\cos\alpha) |
| | + | $$ |
| | + | or |
| | + | $$ |
| | + | \Delta\lambda=\lambda'-\lambda=\frac{h}{m_e c}(1-\cos\alpha) |
| | + | $$ |
| | + | |
| | | #### Answer |
| | | |
| | − | [Insert a concise answer or boxed result] |
| | + | $$ |
| | + | m_{\gamma}=\frac{m}{1+\frac{m}{m_e}(1-\cos\alpha)} |
| | + | $$ |
| | + | $$ |
| | + | m'_e=m_e+m-m_{\gamma} |
| | + | $$ |
