Problem 14.4.16

Solution

Data:

Proton kinetic energy

$E = 10^6\ \text{eV} = 1\ \text{MeV}$, deflection angle $\alpha_p = 0.05\ \text{rad}$

Electron: same kinetic energy

$E = 1\ \text{MeV}$

Rest masses

$m_e c^2 = 0.511\ \text{MeV}, m_p c^2 = 938\ \text{MeV}$

Relationship between deflection and momentum

In a plane capacitor with a uniform transverse electric field, a particle of charge e receives a transverse impulse
$\Delta p_\perp = eE_\text{field}\,\Delta t$ The flight time is $\Delta t \approx L / v$

where L is the length of the plates and v is the initial velocity. For small angles, the deflection is:

$\alpha \approx \frac{\Delta p_\perp}{p} \propto \frac{1}{p v}$

where p is the initial momentum and v is the velocity. The same geometry and field imply$\alpha \propto 1/(p v)$

Calculation of p v for a particle with kinetic energy E

In relativity, the total energy is $E_{\text{total}} = E + mc^2$ The momentum satisfies$(pc)^2 = E_{\text{total}}^2 - (mc^2)^2$ The velocity is$v/c = pc / E_{\text{total}}$

Multiplying:

$p v = \frac{p^2 c^2}{E_{\text{total}}} = \frac{(E + mc^2)^2 - (mc^2)^2}{E + mc^2} = \frac{E(E + 2mc^2)}{E + mc^2}$

Thus, the product pv is expressed in terms of E and the mass m:

$p v = \frac{E(E + 2mc^2)}{E + mc^2}$

Comparison of the angles

Since$\alpha \propto 1/(p v)$ the ratio between the deflection angle of the electron and that of the proton is the inverse of the ratio of their p v products:

$\frac{\alpha_e}{\alpha_p} = \frac{(p v)_p}{(p v)_e}$

Substituting the expressions:

$\frac{\alpha_e}{\alpha_p} = \frac{ \dfrac{E(E + 2m_pc^2)}{E + m_pc^2} }{ \dfrac{E(E + 2m_ec^2)}{E + m_ec^2} } = \frac{E + m_ec^2}{E + 2m_ec^2} \cdot \frac{E + 2m_pc^2}{E + m_pc^2}$

and for small angles $(\alpha_p \ll 1, \alpha_e \ll 1)$ we can approximate$\tan \alpha \approx \alpha$

Numerical calculation

With$E = 1\ \text{MeV}, m_ec^2 = 0.511\ \text{MeV}, m_pc^2 = 938\ \text{MeV}$

$\frac{m_ec^2 + E}{2m_ec^2 + E} = \frac{0.511 + 1}{1.022 + 1} = \frac{1.511}{2.022} \approx 0.7473$

$\frac{2m_pc^2 + E}{m_pc^2 + E} = \frac{1876 + 1}{938 + 1} = \frac{1877}{939} \approx 1.9989 \approx 2.00$.

The product is$0.7473 \times 1.9989 \approx 1.494$.

For $\alpha_p = 0.05\ \text{rad}$:

$\alpha_e \approx 1.494 \times 0.05 \approx 0.0747\ \text{rad} \approx 0.075\ \text{rad}$.