Subfield: Electromagnetism

The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron "biprism" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram).

<img_4450>

The electron "biprism" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the "biprism" is $\ell$, and the distance between the distance between the "biprism" and the screen is $L$.
Context question:
(a)  Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.)
Context answer:
$V(r)=V_{a} \frac{\ln (b / r)}{\ln (b / a)} \text { where } r=\sqrt{x^{2}+z^{2}}$


Context question:
(b)  An incoming electron plane wave with wave vector $k_{z}$ is deflected by the "biprism" due to the $x$-component of the force exerted on the electron. Determine $k_{x}$ the $x$-component of the wave vector due to the "biprism" in terms of the electron charge, $e, v_{\mathrm{z}}, V_{a}, k_{\mathrm{z}}, a$ and $b$, where $e$ and $v_{z}$ are the charge and the $z$-component of the velocity of the electrons $\left(k_{x} \ll k_{z}\right)$. Note that $\vec{k}=\frac{2 \pi \vec{p}}{h}$ where $h$ is the Planck constant.
Context answer:
\boxed{$k_{x} =\frac{e_{0} V_{a} \pi}{\hbar v_{z} \ln \frac{b}{a}}$}


Extra Supplementary Reading Materials:

(c) Before the point $S$, electrons are emitted from a field emission tip and accelerated through a potential $V_{0}$. Determine the wavelength of the electron in terms of the (rest) mass $m$, charge $e$ and $V_{0}$,
Context question:
(i)  assuming relativistic effects can be ignored
Context answer:
\boxed{$\frac{h}{\sqrt{2 m e_{0} V_{0}}}$}


Context question:
(ii)  taking relativistic effects into consideration.
Context answer:
\boxed{$\frac{h}{\sqrt{2 m e_{0} V_{0}\left(1+\frac{e_{0} V_{0}}{2 m_{0} c^{2}}\right)}}$}


Extra Supplementary Reading Materials:

(d) In Tonomura et al experiment,

$$
\begin{array}{ll}
v_{z} & =c / 2, \\
V_{a} & =10 \mathrm{~V}, \\
V_{0} & =50 \mathrm{kV}, \\
a & =0.5 \mu \mathrm{m}, \\
b & =5 \mathrm{~mm}, \\
\ell & =25 \mathrm{~cm}, \\
L & =1.5 \mathrm{~m}, \\
h & =6.6 \times 10^{-34} \mathrm{Js}, \\
\text { electron charge, } e=1.6 \times 10^{-19} \mathrm{C}, \\
\text { mass of electron, } \mathrm{m}_{0}=9.1 \times 10^{-31} \mathrm{~kg}, \\
\text { and the speed of light in vacuo, } c=3 \times 10^{8} \mathrm{~ms}^{-1}
\end{array}
$$
Context question:
(i)  calculate the value of $k_{x}$,
Context answer:
\boxed{$3.46 \times 10^{7} $}


Context question:
(ii)  determine the fringe separation of the interference pattern on the screen,
Context answer:
\boxed{$907$}


Context question:
(iii)  If the electron wave is a spherical wave instead of a plane wave, is the fringe spacing larger, the same or smaller than the fringe spacing calculated in (ii)?
Context answer:
\boxed{larger}


Question: (iv)  In part (c), determine the percentage error in the wavelength of the electron using non-relativistic approximation.

Solution:
Non-relativistic:

$$
\begin{aligned}
\frac{h}{\lambda} & =\sqrt{2 m e_{0} V_{0}} \\
\lambda_{\text {nonrel }} & =\frac{h}{\sqrt{2 m e_{0} V_{0}}} \\
& =5.4697 \times 10^{-12} \mathrm{~m}
\end{aligned}
$$

Relativistic:

$$
\begin{aligned}
\lambda_{\text {rel }} & =\frac{h}{\sqrt{2 m e V_{0}\left(1+\frac{e V_{0}}{2 m_{0} c^{2}}\right)}} \\
& =5.3408 \times 10^{-12} \mathrm{~m}
\end{aligned}
$$

Percentage error:

$$
\begin{aligned}
\text { Error }= & \frac{\lambda_{\text {nonrel }}-\lambda_{\text {rel }}}{\lambda_{\text {rel }}} \\
= & 0.024 \\
& \text { or } 2.4 \text { percent. }
\end{aligned}
$$

Final answer: 2.4 percent