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Chapter 7: Problem 2

By considering momentum and energy conservation in \(\mathrm{e}^{-} p\) elastic scattering from a proton at rest, find an expression for the fractional energy loss of the scattered electron \(\left(E_{1}-E_{3}\right) / E_{1}\) in terms of the scattering angle and the parameter $$ \kappa=\frac{p}{E_{1}+m_{t}} \equiv \frac{\beta \gamma}{\gamma+1} $$

Short Answer

Expert verified
\( \frac{E_1 - E_3}{E_1} = \frac{2 \kappa (1 - \cos\theta)}{1 + 2 \kappa^2 (1 - \cos\theta)} \)

Step by step solution

01

- Define Variables and Given Information

Define initial and final energies of the electron as well as the scattering angle. Initial energy: \[ E_1 \] Final energy: \[ E_3 \] Scattering angle: \[ \theta \] Given parameter: \[ \kappa = \frac{p}{E_1 + m_t} \equiv \frac{\beta \gamma}{\gamma + 1} \]
02

- Conservation Laws

Write down the conservation of momentum and energy equations:For energy conservation:\[ E_1 + m_p = E_3 + E_p \]For momentum conservation (in x and y directions):In x direction:\[ p_1 = p_3 \cos(\theta) + p_p \]In y direction:\[ 0 = p_3 \sin(\theta) \]
03

- Relate Momenta to Energy

Relate the momenta to the energies using the relativistic relations:For electron:\[ p_1 = \sqrt{E_1^2 - m_e^2} \]\[ p_3 = \sqrt{E_3^2 - m_e^2} \]For proton:Provided that proton is initially at rest,\[ p_p = \sqrt{E_p^2 - m_p^2} \]
04

- Express Proton Energy

From energy conservation, solve for proton energy: \[ E_p = E_1 + m_p - E_3 \]
05

- Express in Terms of \( \kappa\)

Using the parameter \( \kappa \), express energies in terms of \( \kappa \):\[ \kappa = \frac{p}{E_1 + m_p} \equiv \frac{p}{E + m}\]
06

- Fractional Energy Loss

From the previously defined relations, express fractional energy loss in terms of the scattering angle and \( \kappa \):The key expression:\[ \frac{E_1 - E_3}{E_1} = \frac{2 \kappa (1 - \cos\theta)}{1 + 2 \kappa^2 (1 - \cos\theta)} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

relativistic energy-momentum relation
The relativistic energy-momentum relation is an essential concept in physics, especially when dealing with particles moving at high velocities close to the speed of light. This relation links the energy of a particle to its momentum and mass. Mathematically, it's expressed as

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Most popular questions from this chapter

For a spherically symmetric charge distribution \(\rho(\mathrm{r})\), where $$ \int \rho(\mathrm{r}) \mathrm{d}^{3} \mathbf{r}=1, $$ show that the form factor can be expressed as $$ \begin{aligned} F\left(\mathbf{q}^{2}\right) &=\frac{4 \pi}{q} \int_{0}^{\infty} r \sin (q r) \rho(r) d r \\ & \simeq 1-\frac{1}{6} q^{2}\left\langle R^{2}\right\rangle+\cdots, \end{aligned} $$ where \(\left\langle R^{2}\right\rangle\) is the mean square charge radius. Hence show that $$ \left\langle R^{2}\right\rangle=-6\left[\frac{d F\left(q^{2}\right)}{d q^{2}}\right]_{q^{2}=0} . $$

The derivation of \((7.8)\) used the algebraic relation $$ (\gamma+1)^{2}\left(1-\kappa^{2}\right)^{2}=4 $$ where $$ \kappa=\frac{\beta \gamma}{\gamma+1} \quad \text { and } \quad\left(1-\beta^{2}\right) \gamma^{2}=1 $$ Show that this holds.
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