$\text{The concentration of electrons in a semiconductor bar varies linearly from}\mathrm{ 2}\times {10}^{17}{\mathrm{cm}}^{-3}\text{at}x=1\mathrm{\mu m}\text{ to}\mathrm{ }$

$1\times {10}^{16}{\mathrm{cm}}^{-3}\text{at}x=4\mathrm{\mu m}.\text{Assume mobility}{\mathrm{ \mu }}_n=1400\frac{{\mathrm{cm}}^2}{V·s}\text{and}{\mathrm{ V}}_T=25\mathrm{mV}.\text{The density of electron diffusion current}$

$\text{(in}\frac{A}{{\mathrm{mm}}^2}\text{) is}\text{\_\_\_\_}\text{(rounded off to two decimal places).}$

The correct answer is 35.47.

1. Identify the Given Parameters:
- Initial electron concentration, $n_1=2\times {10}^{17}{\mathrm{cm}}^{-3}$ at $x_1=1\mathrm{\mu m}=1\times {10}^{���4}\mathrm{cm}$
- Final electron concentration, $n_2=1\times {10}^{16}{\mathrm{cm}}^{-3}$ at $x_2=4\mathrm{\mu m}=4\times {10}^{-4}\mathrm{cm}$
- Electron mobility, ${\mu }_n=1400{\mathrm{cm}}^2/(V·s)$
- Thermal voltage, $V_T=25\mathrm{mV}=0.025V$
- Elementary charge, $q\approx 1.6\times {10}^{-19}C$

2. Calculate the Electron Diffusion Coefficient (D_n):
Using Einstein's relation:
$D_n={\mu }_n·V_T$
$D_n=1400\times 0.025=35{\mathrm{cm}}^2/s$

3. Calculate the Concentration Gradient (dn/dx):
Since the concentration varies linearly:
$\frac{\mathrm{dn}}{\mathrm{dx}}=\frac{n_2-n_1}{x_2-x_1}$
$\frac{\mathrm{dn}}{\mathrm{dx}}=\frac{1\times {10}^{16}-2\times {10}^{17}}{(4-1)\times {10}^{-4}}=\frac{-1.9\times {10}^{17}}{3\times {10}^{-4}}\approx -6.333\times {10}^{20}{\mathrm{cm}}^{-4}$

4. Calculate the Magnitude of the Electron Diffusion Current Density (J_n):
The equation for electron diffusion current density is:
$J_n=q·D_n·\left|\frac{\mathrm{dn}}{\mathrm{dx}}\right|$
$J_n=(1.6\times {10}^{-19})\times 35\times (6.333\times {10}^{20})$
$J_n=5.6\times {10}^{-18}\times 6.333\times {10}^{20}=3546.67A/{\mathrm{cm}}^2$

5. Convert to A/mm²:
Since $1{\mathrm{cm}}^2=100{\mathrm{mm}}^2$:
$J_n=\frac{3546.67}{100}=35.4667A/{\mathrm{mm}}^2$
Rounding off to two decimal places, we get 35.47.