The energy band diagram and the electron density profile n(x) in a semiconductor are shown in the figures. Assume that \({\rm{n}}\left( {\rm{x}} \right) = 10_{\rm{e}}^{15}\left( {\frac{{{\rm{qax}}}}{{{\rm{kT}}}}} \right){\rm{c}}{{\rm{m}}^{ - 3}}\), with \({\rm{\alpha }} = 0.1{\rm{V}}/{\rm{cm}}\) and x expressed in cm. Given \(\frac{{{\rm{kT}}}}{{\rm{q}}} = 0.026{\rm{V}},{\rm{\;}}{{\rm{D}}_{\rm{n}}} = 36{\rm{c}}{{\rm{m}}^2}{{\rm{s}}^{ - 1}},{\rm{\;and}}\frac{{\rm{D}}}{{\rm{\mu }}} = \frac{{{\rm{kT}}}}{{\rm{q}}}\). The electron current density (in A/cm2) at x = 0 is

The energy band diagram and the electron density profile n(x) in a sem

1.	The energy band diagram and the electron density profile n(x) in a semiconductor are shown in the figures. Assume that \({\rm{n}}\left( {\rm{x}} \right) = 10_{\rm{e}}^{15}\left( {\frac{{{\rm{qax}}}}{{{\rm{kT}}}}} \right){\rm{c}}{{\rm{m}}^{ - 3}}\), with \({\rm{\alpha }} = 0.1{\rm{V}}/{\rm{cm}}\) and x expressed in cm. Given \(\frac{{{\rm{kT}}}}{{\rm{q}}} = 0.026{\rm{V}},{\rm{\;}}{{\rm{D}}_{\rm{n}}} = 36{\rm{c}}{{\rm{m}}^2}{{\rm{s}}^{ - 1}},{\rm{\;and}}\frac{{\rm{D}}}{{\rm{\mu }}} = \frac{{{\rm{kT}}}}{{\rm{q}}}\). The electron current density (in A/cm2) at x = 0 is
A.	\(- 4.4 \times {10^{ - 2}}\)
B.	\(- 2.2 \times {10^{ - 2}}\)
C.	\(0\)
D.	\(2.2 \times {10^{ - 2}}\)
Answer» D. \(2.2 \times {10^{ - 2}}\)