$$\begin{array}{rl}& \frac{dm}{dt}=-p_{g}m+p_3{m}_{e2}\\ & \frac{d{m}_{e2}}{dt}=p_{g}m-p_2{m}_{e2}-p_3{m}_{e2}=p_{g}m-(p_2+p_3)m_{e2}\end{array}$$ [mathjax]

$$\frac{d}{dt}\left[\begin{array}{c}m\\ m_{e2}\end{array}\right]=\left[\begin{array}{cc}-p_g& p_3\\ p_g& -(p_2+p_3)\end{array}\right]\left[\begin{array}{c}m\\ m_{e2}\end{array}\right]$$

方程的解为$\left[\begin{array}{c}m\\ m_{e2}\end{array}\right]=c_1{e}^{{\lambda }_{1}t}{\mathbf{v}}_{\mathbf{1}}+c_2{e}^{{\lambda }_{2}t}{\mathbf{v}}_{\mathbf{2}}$

$$\begin{array}{rl}& {\lambda }_1+{\lambda }_2=-(p_2+p_3+p_g)\\ & {\lambda }_1{\lambda }_2=p_2{p}_g\end{array}$$

$$p_2=s\exp (\frac{-\mathrm{\Delta }E}{kT})$$

$$F(R)=\frac{(1-R_{\mathrm{\infty }}{)}^2}{2R_{\mathrm{\infty }}}\propto \frac{\epsilon c}{s}$$

$$F(R)=\frac{K}{S}=\frac{{(1-R_{\mathrm{\infty }})}^2}{2R_{\mathrm{\infty }}}\propto m$$

$$R_{\mathrm{\infty }}=\frac{I_r}{I_{in}}$$

$$m(t)=cf(I_r(t),I_{in})$$

$$F(R)=\frac{{(1-R_{\mathrm{\infty }})}^2}{2R_{\mathrm{\infty }}}\propto \frac{\sigma c}{s}$$

$${\lambda }_1{\lambda }_2=p_2{p}_g=p_g\mathrm{sexp}\left(\frac{-\mathrm{\Delta }E}{kT}\right)$$

$$g(T)=p_g\mathrm{sexp}\left(\frac{-\mathrm{\Delta }E}{kT}\right)$$

$$\ln g(T)=\frac{-\mathrm{\Delta }E}{kT}+\ln p_{g}s$$

$$\frac{1}{kT}$$