Problem 7.5

Enter all the constants in consistent units (or let Mathematica do all the unit conversions).

k = Quantity[1, "BoltzmannConstant"];
h = Quantity[1, "PlanckConstant"];
m = Quantity[1, "ElectronMass"];
Ie = Quantity[0.044, "Electronvolts"] (* Ionization energy *);
nd = Quantity[(10^17), 1/"Centimeters"^3] (* Donor density *);

Enter the dimensionless helper function.  Assume we will enter temperatures in Kelvin.

y[T_] := nd (((h^3)/(((2 π m k Quantity[T, "Kelvins"]))^(3/2))))Exp[Ie/(k Quantity[T, "Kelvins"])]

x[T_] := (1/(2 y[T]))(Sqrt[1 + 4 y[T]] - 1)

x[T]

Plot[x[T], {T, 5, 300}, LabelStyle->Larger, AxesLabel -> {"T (K)", "(Subscript[N, c])/(Subscript[N, d])"}, ImageSize -> Scaled[0.8], PlotLabel->"Fraction of Ionized Donors"]

At room temperature, nearly all of the donors will be ionized.  The temperature at which half of them will be ionized is about 122 K:

FindRoot[x[T] == 0.5, {T, 50}]

Further explorations: Chemical Potential

The chemical potential for Nd donated electrons is

lQ[T_] := (h/(Sqrt[2 π m k T]))

UnitConvert[lQ[Quantity[300, "Kelvins"]], "Nanometers"]

vQ[T_] := ((lQ[T])^3)

(* Assuming Nc = Nd (i.e. every donor is ionized *)
μ[T_]  := UnitConvert[-k Quantity[T, "Kelvins"]  Log[(2/(nd vQ[Quantity[T, "Kelvins"]]))], "Electronvolts"]

Plot[{-Ie,μ[T]}, {T, 50, 200}]