D.29 The thermionic emis­sion equa­tion

This note de­rives the thermionic emis­sion equa­tion for a typ­i­cal metal fol­low­ing [41, p. 364ff]. The de­riva­tion is semi-clas­si­cal.

To sim­plify the analy­sis, it will be as­sumed that the rel­e­vant elec­trons in the in­te­rior of the metal can be mod­eled as a free-elec­tron gas. In other words, it will be as­sumed that in the in­te­rior of the metal the forces from sur­round­ing par­ti­cles come from all di­rec­tions and so tend to av­er­age out.

(The free-elec­tron gas as­sump­tion is typ­i­cally qual­i­ta­tively rea­son­able for the va­lence elec­trons of in­ter­est if you de­fine the zero of the ki­netic en­ergy of the gas to be at the bot­tom of the con­duc­tion band. You can also re­duce er­rors by re­plac­ing the true mass of the elec­tron by some suit­able ef­fec­tive mass. But the zero of the en­ergy drops out in the fi­nal ex­pres­sion, and the ef­fec­tive mass of typ­i­cal sim­ple met­als is not greatly dif­fer­ent from the true mass. See chap­ter 6.22.3 for more on these is­sues.)

As­sume that the sur­face through which the elec­trons es­cape is nor­mal to the $x$-​di­rec­tion. Then the clas­si­cal ex­pres­sion for the cur­rent of es­cap­ing elec­trons is

\begin{displaymath}
j = \rho e v_x
\end{displaymath}

where $\rho$ is the num­ber of elec­trons per unit vol­ume that is ca­pa­ble of es­cap­ing and $v_x$ is their ve­loc­ity in the $x$-​di­rec­tion. Note that the cur­rent above is sup­posed to be the cur­rent in­side the metal of the elec­trons that will es­cape.

An elec­tron can only es­cape if its en­ergy ${\vphantom' E}^{\rm p}$ ex­ceeds

\begin{displaymath}
{\vphantom' E}^{\rm p}_{\rm {esc}}=\mu+e\varphi_{\rm {w}}
\end{displaymath}

where $\mu$ is the Fermi level, be­cause the work func­tion $\varphi_{\rm {w}}$ is de­fined that way. The num­ber of elec­trons per unit vol­ume in an en­ergy range ${\rm d}{\vphantom' E}^{\rm p}$ above ${\vphantom' E}^{\rm p}_{\rm {esc}}$ can be found as

\begin{displaymath}
e^{-(e\varphi_{\rm {w}}+{\vphantom' E}^{\rm p}-{\vphantom' ...
... \sqrt{{\vphantom' E}^{\rm p}} {\,\rm d}{\vphantom' E}^{\rm p}
\end{displaymath}

That is be­cause the ini­tial ex­po­nen­tial is a rewrit­ten Maxwell-Boltz­mann dis­tri­b­u­tion (6.21) that gives the num­ber of elec­trons per state, while the re­main­der is the num­ber of states in the en­ergy range ac­cord­ing to the den­sity of states (6.6).

Nor­mally, the typ­i­cal ther­mal en­ergy ${k_{\rm B}}T$ is very small com­pared to the min­i­mum en­ergy $e\varphi_{\rm {w}}$ above the Fermi level needed to es­cape. Then the ex­po­nen­tial of the Maxwell-Boltz­mann dis­tri­b­u­tion is very small. That makes the amount of elec­trons with suf­fi­cient en­ergy to es­cape very small. In ad­di­tion, with in­creas­ing en­ergy above ${\vphantom' E}^{\rm p}_{\rm {esc}}$ the amount of elec­trons very quickly be­comes much smaller still. There­fore only a very small range of en­er­gies above the min­i­mum en­ergy ${\vphantom' E}^{\rm p}_{\rm {esc}}$ gives a con­tri­bu­tion.

Fur­ther, even if an elec­tron has in prin­ci­ple suf­fi­cient en­ergy to es­cape, it can only do so if enough of its mo­men­tum is in the $x$-​di­rec­tion. Only mo­men­tum that is in the $x$-​di­rec­tion can be used to over­come the nu­clei that pull it back to­wards the sur­face when it tries to es­cape. Mo­men­tum in the other two di­rec­tions only pro­duces mo­tion par­al­lel to the sur­face. So only a frac­tion, call it $f_{\rm {esc}}$, of the elec­trons that have in prin­ci­ple enough en­ergy to es­cape can ac­tu­ally do so. A bit of geom­e­try shows how much. All pos­si­ble end points of the mo­men­tum vec­tors with a mag­ni­tude $p$ form a spher­i­cal sur­face with area $4{\pi}p^2$. But only a small cir­cle on that sur­face around the $x$-​axis, with an ap­prox­i­mate ra­dius of $\sqrt{p^2-p_{\rm {esc}}^2}$, has enough $x$-​mo­men­tum for the elec­tron to es­cape, so

\begin{displaymath}
f_{\rm {esc}} \approx \frac{\pi\sqrt{p^2-p_{\rm {esc}}^2}^2...
...- {\vphantom' E}^{\rm p}_{\rm {esc}}}{4{\vphantom' E}^{\rm p}}
\end{displaymath}

where the fi­nal equal­ity ap­plies since the ki­netic en­ergy is pro­por­tional to the square mo­men­tum.

Since the ve­loc­ity for the es­cap­ing elec­trons is mostly in the $x$-​di­rec­tion, ${\vphantom' E}^{\rm p}$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $\frac12{m_{\rm e}}v_x^2$, which can be used to ex­press $v_x$ in terms of en­ergy.

Putting it all to­gether, the cur­rent den­sity be­comes

\begin{displaymath}
j =
\int_{{\vphantom' E}^{\rm p}={\vphantom' E}^{\rm p}_{\...
...{\rm p}}{m_{\rm e}}\right)^{1/2} {\rm d}{\vphantom' E}^{\rm p}
\end{displaymath}

Rewrit­ing in terms of a new in­te­gra­tion vari­able $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $({\vphantom' E}^{\rm p}-{\vphantom' E}^{\rm p}_{\rm {esc}})$$\raisebox{.5pt}{$/$}$${k_{\rm B}}T$ gives the thermionic emis­sion equa­tion.

If an ex­ter­nal elec­tric field ${\cal E}_{\rm {ext}}$ helps the elec­trons es­cape, it low­ers the en­ergy that the elec­trons need to do so. Con­sider the po­ten­tial en­ergy in the later stages of es­cape, at first still with­out the ad­di­tional elec­tric field. When the elec­tron looks back at the metal sur­face that it is es­cap­ing from, it sees a positron mir­ror im­age of it­self in­side the metal. Of course, there is not re­ally a positron in­side the metal; re­arrange­ment of the sur­face elec­trons of the metal cre­ate this il­lu­sion. The sur­face elec­trons re­arrange them­selves to make the to­tal com­po­nent of the elec­tric field in the di­rec­tion par­al­lel to the sur­face zero. In­deed, they have to keep mov­ing un­til they do so, since the metal has neg­li­gi­ble elec­tri­cal re­sis­tance in the di­rec­tion par­al­lel to the sur­face. Now it just so hap­pens that a positron mir­ror im­age of the elec­tron has ex­actly the same ef­fect as this re­arrange­ment. The es­cap­ing elec­tron pushes the sur­face elec­trons away from it­self; that force has a re­pul­sive com­po­nent along the sur­face. The positron mir­ror im­age how­ever at­tracts the sur­face elec­trons to­wards it­self, ex­actly can­celling the com­po­nent of force along the sur­face ex­erted by the es­cap­ing elec­tron.

The bot­tom line is that it seems to the es­cap­ing elec­tron that it is pulled back not by sur­face charges, but by a positron mir­ror im­age of it­self. There­fore, in­clud­ing now an ad­di­tional ex­ter­nal elec­tri­cal field, the to­tal po­ten­tial in the later stages of es­cape is:

\begin{displaymath}
V = - \frac{e^2}{16\pi\epsilon_0 d} - e {\cal E}_{\rm ext} d + \mbox{constant}
\end{displaymath}

where $d$ is the dis­tance from the sur­face. The first term is the at­tract­ing force due to the positron im­age, while the sec­ond is due to the ex­ter­nal elec­tric field. The con­stant de­pends on where the zero of en­ergy is de­fined. Note that only half the en­ergy of at­trac­tion be­tween the elec­tron and the positron im­age should be as­signed to the elec­tron; the other half can be thought of as work on the im­age. If that is con­fus­ing, just write down the force on the elec­tron and in­te­grate it to find its po­ten­tial en­ergy.

If there is no ex­ter­nal field, the max­i­mum po­ten­tial en­ergy that the elec­tron must achieve oc­curs at in­fi­nite dis­tance $d$ from the metal sur­face. If there is an elec­tric field, it low­ers the max­i­mum po­ten­tial en­ergy, and it now oc­curs some­what closer to the sur­face. Set­ting the de­riv­a­tive of $V$ with re­spect to $d$ to zero to iden­tify the max­i­mum, and then eval­u­at­ing $V$ at that lo­ca­tion shows that the ex­ter­nal field low­ers the max­i­mum po­ten­tial en­ergy that must be achieved to es­cape by $\sqrt{e^3{\cal E}/4\pi\epsilon_0}$.