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\exnumber{6030}
\heading{Thermionic emission of electrons from a metal} 
\auname{Noam Wunch (part 1) and Gideon Ilani (parts 2-3)}


{\bf The problem:}

\Dn

A spherical piece of metal (”cathode”), that has radius R and temperature T , is placed inside a vacuum tube. A second metallic piece (”anode”) is used to collect the electrons that are emitted from the cathode. The effective temperature of the anode is zero. The cathode has a work function $W$, while the anode has work function $W^{\prime}$. The depth of the potential that holds the electrons inside the cathode, aka the potential floor, is $V_{0}$. 

\Dn

(1) Write an integral expression for the saturation current $I_{s}$ that would be measured when the bias voltage is very large. 

(1a) Show that $V_{0}$ does not appear in the final result: the outcome of the calculation is the same for sections that are close to the surface or deep in the metal.

(1b) Calculate the integral using the Boltzmann approximation. Specify the range of temperatures for which the approximation is valid. 

(2) Using the result of the previous item write an estimate for the current if a reverse (stopping) voltage $V_{\mathrm{battery}}$ is applied. Explain whether $W$ or $W^{\prime}$ is relevant. 

(2a) Explain the relation to the analysis of the stopping voltage in the photoelectric effect. 

(3) Assume that the cathode is detached and left alone in free space. Calculate the charge $Q\left(t\right)$ of the cathode as a function of time assuming that $Q\left(0\right)=0$. 

(3a) Explain the limitations of the result that you have obtained.

\Dn\Dn

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{\bf The solution:}
\Dn

(1) 
With the potential floor at $E=0$, an electron on the cathode surface has energy
\begin{equation}
E = \frac{1}{2}mv^2 = \frac{1}{2}m\vec{v}_{\parallel}^2 + \frac{1}{2}mv_{\perp}^2.
\end{equation}
Where $\vec{v}_{\parallel}$ and $v_{\perp}$ are the electron velocities in the direction parallel and perpendicular to the surface respectively. In order to escape, the electron's kinetic energy in the direction perpendicular to the surface should be greater than the potential barrier
\begin{equation}
\frac{1}{2}mv_{\perp}^2 > eV_0 \longrightarrow v_{\mathrm{\perp,min}} = \sqrt{\frac{2eV_{0}}{m}}. 
\end{equation}
For a large (repelling) bias voltage, all escaping electrons reach the anode and contribute to the current. Hence we will integrate over all electrons with $v_{\perp}>v_{\mathrm{\perp,min}}$.
\begin{figure}[h!]
\begin{center} \putgraph{picture1.png}\caption{System diagram} \end{center}
\end{figure}
\Dn

The current flux in a direction perpendicular to a surface is given by:
\begin{equation}
J = \int\frac{f(\vec{v})d^3\vec{v}}{V}(\vec{v}\cdot\hat{n}) = \int\frac{f(\vec{v})d^3\vec{v}}{V}v_{\perp}
\end{equation}
Where $f(\vec{v})$ is the velocity distribution, and is given by:
\begin{equation}
f(\vec{v})=2 \times V \times \left(\frac{m}{2 \pi}\right)^{3}f\left(\frac{1}{2}mv^2-\mu\right)
\end{equation}
In the above equation, $f$ is Fermi occupation and the preceding factor of 2 accounts for spin degeneracy. By definition of W, and as depicted in Fig(1), $\mu = eV_0 -W$. All together we have:
\begin{equation}
J = 2 \left(\frac{m}{2\pi}\right)^3 \int_{-\infty}^{\infty}d^2v_{\parallel} \int_{v_{\mathrm{\perp,min}}}^{\infty}f\left(\frac{1}{2}m\vec{v}_{\parallel}^2 + \frac{1}{2}mv_{\perp}^2-eV_0+W\right)v_{\perp}dv_{\perp}
\end{equation}

%%%%%%%%%%%%%%%
\Dn

(1a)
In order to show this result is independent of $V_0$ we define $\epsilon=\frac{1}{2}mv_{\perp}^2$. Eq(5) can be rewritten as:
\begin{equation}
I_s = 4\pi R^2 \times J =4\pi R^2 \times 2 \left(\frac{m}{2\pi}\right)^3\int_{-\infty}^{\infty}d^2v_{\parallel} \int_{eV_0}^{\infty}f\left(\frac{1}{2}m\vec{v}_{\parallel}^2 + \epsilon -eV_0 +W\right)\frac{d\epsilon}{m}
\end{equation}
Where we used $\frac{1}{2}mv_{\mathrm{\perp,min}}^2=eV_0$, as per Eq(2). It is evident, by shifting $\epsilon$ to $\tilde{\epsilon}=\epsilon-eV_0$, that the equation above is independent of $V_0$.
\begin{equation}
I_s = 4\pi R^2 \times J =4\pi R^2 \times 2 \left(\frac{m}{2\pi}\right)^3\int_{-\infty}^{\infty}d^2v_{\parallel} \int_{0}^{\infty}f\left(\frac{1}{2}m\vec{v}_{\parallel}^2 + \tilde{\epsilon} +W\right)\frac{d\tilde{\epsilon}}{m}
\end{equation}
Physically this is due to detailed balance: In equilibrium, the net current is zero, so if there is a current into the cathode the current out of it will be determined by the demand that it will cancel the current into it. The current into the cathode from the vacuum is clearly not related to the cathode potential floor, so this is true also for the current out of the cathode.


%%%%%%%%%%%%%% 4\pi R^2 \times 2\pi \left(\frac{m}{2\pi}\right)^3 \left(\frac{2T}{m}\right)^2 e^{-\beta W} \times \int_{0}^{\infty}x^3e^{-\beta x^2)}dx =
\Dn

(1b)
In the Boltzmann approximation the velocity distribution becomes:
\begin{equation}
f(v)= e^{-\beta\left(\frac{1}{2}mv^2-\mu\right)}=e^{-\beta\left(\frac{1}{2}m\vec{v}_{\parallel}^2 + \tilde{\epsilon} +W\right)}
\end{equation}
This approximation holds whenever the occupation $f$ is small. If it holds for the electrons with minimal energy to escape it holds for all higher levels. Accordingly, the condition is $e^{-\beta(eV_0- \mu)}<<1$ which implies $T<<eV_0-\mu$ or $T<<W$. After plugging Eq(8) into Eq(7) and changing to dimensionless variables the integral can be rewritten as:
\begin{equation}
I_s = 4\pi R^2 \times 2 \left(\frac{m}{2\pi}\right)^3e^{-\frac{W}{T}} \times \left(\frac{2T}{m}\right)^2 \int_{-\infty}^{\infty}e^{-y^2}d^2y \int_{0}^{\infty}e^{-x^2}xdx
\end{equation}
The first integral gives a factor of $\pi$ and the second a factor of $\frac{1}{2}$, finally:
\begin{equation}
I_s = 4\pi R^2 \times \frac{m}{2\pi^2} T^2 e^{-\frac{W}{T}}
\end{equation}
%%%%%%%%%%%%%
\Dn

(2)
In the above relation we see that the thermionic emission here is proportional to $T^2 e^{-\beta W}$. If there is no stopping voltage the energetic barrier an electron has to pass is $W$. When there is stopping voltage the energetic barrier becomes 
\begin{equation}
\mu_{\mathrm{anode}}-\mu_{\mathrm{cathode}}+W^{\prime}=W^{\prime}+eV_{\mathrm{battery}},
\end{equation}
 so we expect that the expression for the current will be:
\begin{equation}
I\propto T^{2}e^{-\beta\left(W^{\prime}+eV_{\mathrm{battery}}\right)}
\end{equation}
We used $\propto$ because our argument does not account for the geometric prefactors.

%%%%%%%%%%%%%
\Dn

(2a)
In the photoelectric effect the cathode isn't heated so one assumes zero temperature Fermi occupation for the cathode also. That means there is no thermionic emission. The electrons are excited out of the cathode by photons with a fixed energy. These photons need to be with enough energy for the electrons to pass the energetic barrier, which is again $W^{\prime}+eV_{\mathrm{battery}}$. So if we measure the  voltage needed to stop the current for a given photons energy, we know that at that voltage we have:
\begin{equation}
E_{\mathrm{photon}}=W^{\prime}+eV_{\mathrm{battery}} \implies W^{\prime}=E_{\mathrm{photon}}-eV_{\mathrm{battery}}
\end{equation}
So we can calculate the anode work function.
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\Dn

(3)
We can treat the cathode as a capacitor. The charge will be $Q=VC$, and the current: $I=C\frac{d}{dt}V$. Here $C=\frac{1}{R}$, and $V$ are the capacitance and voltage of a charged sphere in vacuum with respect to infinity. On the other hand we know from item (2) that when there is a voltage difference the current is $\sim I_{0}e^{-\beta eV}$ , where $I_{0}$ is independent of $V$. So we need to solve:
\begin{equation}
\frac{d}{dt}V=\frac{I_{0}}{C}e^{-\beta eV}
\end{equation}
given that $Q\left(t=0\right)=0$:
\begin{equation}
e^{\beta eV}dV=\frac{I_{0}}{C}dt\implies\frac{T}{e}e^{\beta eV}=\frac{I_{0}}{C}t+\frac{T}{e}
\end{equation}
\begin{equation}
V=\frac{T}{e}\ln\left(\frac{e}{T}\frac{I_{0}}{C}t+1\right)\implies Q \left(t\right)=\frac{CT}{e}\ln\left(\frac{e}{T}\frac{I_{0}}{C}t+1\right)
\end{equation}

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\Dn

(3a)
The result is limited because it is valid only when the Boltzmann approximation is valid, and it does not take into consideration finite number of electrons.

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