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The energy band diagram and the electron density profile $n(x)$ in a semiconductor are shown in the figures. Assume that $n(x) = 10^{15}e^{\left(\dfrac{q\alpha x}{kT}\right)}\:cm^{-3} ,$ with $\alpha = 0.1\:V/cm$ and $x$ expressed in cm. Given $\dfrac{kT}{q} = 0.026\;V,D_{n} = 36\:cm^{2}s^{-1},$  and $\dfrac{D}{\mu} = \dfrac{kT}{q}.$ The electron current density $(\text{in}\: A/cm^{2})$ at $x=0$ is

  1. $-4.4\times 10^{-2}$
  2. $-2.2\times 10^{-2}$
  3. $0$
  4. $2.2\times 10^{-2}$
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