Photoemission Current - Photoelectrochemistry

Question

This book on photoelectrochemistry calculates the emission current from a semiconductor as a function of changing frequency (see Equation 1.13, but reproduced below). The expression for the photoemission current $I$ directed perpendicularly to emitter's surface is given by:

$$ I=e \ \rho_{0} \int j_x \left[e^{\left(E_i-\mu\right) / k T}+1\right]^{-1} d E_{\mathrm{i}} d \mathbf{p}_{\|} $$

where the $E_i$ and $\mathbf{p}_{\|}=\left\{p_y, p_z\right\}$ variables are the energy and parallel (to the surface) components of the momentum of the initial electrons in the metal, respectively. The density of states, $\rho_{0}$ is assumed constant. The electron current, $j_x$ is dependent on energy $E = p^{2}/2m$.

The limits of integration with respect to $E$ and $p_{\|}^2$ are obtained from the law of conservation of energy:

$$ p=\sqrt{2 m (E_{i}+\hbar \omega)-p_{\|}^2}, $$

where $p$ and $m$ denote the $x$-component of the momentum and the mass, respectively, of the emitted electron. Since the partial photocurrent $J$ differs from zero only for real values of $p$, the integration must extend over all values of $E$ and $p_{\|}$which satisfy the condition

$$ \begin{gathered} 2 m(E_{i}+\hbar \omega) \geqslant p_{\|}^2 \text { or } \\ 0 \leqslant p_{\|}^2 \leqslant 2 m(E_{i}+\hbar \omega),-\hbar \omega<E<\infty . \end{gathered} $$

$p$ depends only on $\mathbf{p}_{\|}{ }^2=\left|\mathbf{p}_{\|}\right|^2$. Therefore, the integral can be immediately integrated once to obtain $d \mathbf{p}_{\|}=2 \pi\left|\mathbf{p}_{\|}\right| d\left|\mathbf{p}_{\|}\right|$.

Introducing, in place of $E_i$ and $\left|\mathbf{p}_{\|}\right|$, the new variable $E \equiv p^2 / 2 m$, we can integrate over $E_i$. Thee final result, the absolute value of $I$ is given by:

$$ I=2 \pi e \rho_0 m k T \int_0^{\infty} j_x(\sqrt{2 m E}) \ln \left[1+e^{(\mu+\hbar w-E) / k T}\right] d E $$

I do not understand this step. I do not see or work out the substitution for the exponential.

There is a reference for this step, a Russian textbook - Brodskii and Gurevich, 1973 - Brodskii, A. M., and Gurevich, Yu. Va. (1973). Theory of Electronic Emission from Metals. Nauka, Moscow. But it isn't available anywhere I can find. Apparently page 7 has the solution.

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An Attempt:

I think - but I'm not sure since I don't get the correct result, but, taking the first expression,

$$ I=e \rho_0 \int j_x\left[e^{\left(E_i-\mu\right) / k T}+1\right]^{-1} d E_{\mathrm{i}} d \mathbf{p}_{\|} $$

using $k_{y}^2+k_{z}^2=p_{\|}^2 $ and the fact that $ \int_0^{\infty} \left(1+e^x / b\right)^{-1} \mathrm{d} x = \ln (1+b)$.

Hence

$$ \int j_x\left[e^{\left(E_i-\mu\right) / k T}+1\right]^{-1} d \mathbf{p}_{\|}\\ = \int j_x\left[e^{\left(\left[\hbar^2\left(k_x^2+k_y^2+ k_z^2\right) /(2 m)\right]-\mu\right) / k T}+1\right]^{-1} d \mathbf{p}_{\|}\\ = \pi \int_{0}^{\infty} j_x\left[e^{\left(\left[\hbar^2\left(p_{\|}^2+ k_x^2\right) /(2 m)\right]-\mu\right) / k T}+1\right]^{-1} dp_{\|}^{2}\\ = j_x \frac{2 \pi m k_{\mathrm{B}} T}{\hbar^2} \ln \left[1+e^{\frac{\mu-\hbar^2 k_{z}^2 /(2 m)}{k_{\mathrm{B}} T}}\right]\\ $$

and so the integral becomes

$$ I=e \rho_0 \frac{2 \pi m k_{\mathrm{B}} T}{\hbar^2} \int j_x \ln \left[1+e^{\frac{\mu-\hbar^2 k_{x}^2 /(2 m)}{k_{\mathrm{B}} T}}\right] d E_{\mathrm{i}} $$

$\hbar^2 k_{x}^2 /(2 m)$ is now the energy of the remaining component.

Answer 1

I've found a copy of the textbook (in Russian) from Princeton University library, who have very kindly offered to send it to me. I will learn russian, then read the textbook and then report back.

But, until then, I have something that resembles a solution, but has some (obvious) inconsistencies.

Starting with

$$ I=\int j\left[e^{(E-\mu) / k_B T}-1\right]^{-1} \rho_0 d E d \vec{{p}_{\|}} $$

Integrating $d \vec{{p}_{\|}}$ over all angles

$$ \begin{aligned} & \int ^\theta d \vec{{p}_{\|}} \Rightarrow 2 \pi\left|{p}_{\|}\right| d\left|{p}_{\|}\right|=d \vec{{p}_{\|}} \\ \Rightarrow & I=2 \pi \int j\left|\vec{{p}_{\|}}\right|\left[e^{(E-\mu) / k_B T}-1\right]^{-1} \rho_0 d E d\left|{p}_{\|}\right| \\ \end{aligned} $$

Suppose that ${p}_{\|}$ is always positive $$ \left|p_{11}\right|={p}_{\|} \\ \Rightarrow I=2 \pi \int j {p}_{\|}\left[e^{(E-\mu) / k_B T}\right]^{-1} \rho_0 d E d {p}_{\|}$$

Now substitute

$$ \begin{gathered} E=\frac{1}{2 m}\left({p}_{\|}^2+p_z^2\right)-\hbar \nu \\ \Rightarrow I=2 \pi \int j {p}_{\|}\left[e^{\left(\frac{1}{2 m}\left({p}_{\|}^2+p_z^2\right)-\hbar \nu-r\right) / k_B^T}{ }-1\right]^{-1} \rho_0 d E d {p}_{\|} \end{gathered} $$

Now integrate over $d{p}_{\|}$ (from $0$ to $\infty$ ?!?!), gives

$$ \begin{aligned} & =2 \pi \rho_0 \frac{k_B T_m}{\hbar^2} \int j \ln \left[1+e^{\left(\mu+\hbar v-p_{z}^{2} / 2 m\right) / h_B T}\right] d E \end{aligned} $$

Now changing integration variables to $p_{z}$, $$ \begin{aligned} & E=\frac{1}{2 m}\left({p}_{\|}^2+p_z^2\right)-\hbar \nu \\ & \frac{d E}{\partial p_z}=\frac{p_z}{m} \Rightarrow d E=\frac{p_z}{m} d p_z \\ & =2 \pi \rho_0 \frac{k_B T m}{\hbar^2} \int j \ln \left[1+e^{\left.\left(\mu+\hbar r-p_z^2 / 2 m\right) / k_B T\right]} \frac{p_z}{m} d p_z\right. \\ & =2 \pi \rho_0 \frac{k_B T}{\hbar} \int_0^{\infty} j \ln \left[1+e^{\left(\mu+\hbar r-p_z^2 / 2 m\right) / h_B T}\right] p_{z} d p_z \end{aligned} $$

Now, this looks very much like the above expression, apart from if you change to $p_{z}\rightarrow E or E_{z}$, you'll lose the $p_{z}$ and there is no way to get the $\sqrt{E_{z}2m}$ in the integral.