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Show that the resistivity of (1 pure germanium (Ge) at room temperature is 0.45 ohm. meter. Given u_n=38 m2/V.s , µ_p=O.18 m2/V.s, ni= 2.5*10^19 m^-3, atomic weight=72.6, density=5.32*1O^3 Kg/m3 2) If 1 atom phosphorus added to every 10^8 of Ge atoms, show the resistivity becomes 0.037 ohm. meter

Show that the resistivity of (1 pure germanium (Ge) at room temperature is 0.45 ohm. meter. Given u_n=38 m2/V.s , µ_p=O.18 m2/V.s, ni= 2.5*10^19 m^-3, atomic weight=72.6, density=5.32*1O^3 Kg/m3 2) If 1 atom phosphorus added to every 10^8 of Ge atoms, show the resistivity becomes 0.037 ohm. meter

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Show that the resistivity of (1 pure germanium (Ge) at room temperature is 0.45 ohm. meter. Given u_n=38 m2/V.s, u_p=D0.18 m2/V.s, ni= 2.5*10^19 m^-3, atomic weight=72.6, density=5.32*10^3 Kg/m3 2) If 1 atom phosphorus added to every 10^8 of Ge atoms, show the resistivity becomes 0.037 ohm. meter
Transcribed Image Text:Show that the resistivity of (1 pure germanium (Ge) at room temperature is 0.45 ohm. meter. Given u_n=38 m2/V.s, u_p=D0.18 m2/V.s, ni= 2.5*10^19 m^-3, atomic weight=72.6, density=5.32*10^3 Kg/m3 2) If 1 atom phosphorus added to every 10^8 of Ge atoms, show the resistivity becomes 0.037 ohm. meter
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