The intrinsic concentration of electrons and holes, n_i, in Germanium (Ge) at ro

Question

The intrinsic concentration of electrons and holes, n_i, in Germanium (Ge) at room temperature is approximately 1 times 10^19 m^-3. Given the density and atomic mass of greaterthanorequalto to be 5.32 g/cm^3 and 72.61 g/mol, determine the ratio of ionized to unionized greaterthanorequalto atoms in the lattice. Assume that each greaterthanorequalto atom can be ionized only once (only one of its 4 covalent bonds can be broken ... greaterthanorequalto has a diamond cubic crystal structure). Then, using the value of n_i you should be able to estimate the number of broken covalent bonds per unit volume. Remember that each broken bond leads directly to one free electron-hole pair. Finally, compute the ratio of broken covalent bonds per unit volume to total covalent bonds per unit volume. A. 8.21 times 10^-6 B. 2.00 times 10^-19 C. 2.27 times 10^-10 D. 2.27 times 10^-4 E. 4.54 times 10^-14 F. 4.54 times 10^-4 G. 2.00 times 10^-13

Explanation / Answer

Atomic mass of Ge = 72.61 g/mol

Density = 5.32 g/cm3 = 5.32*106 g/m3

volume of 1 mol of Ge = mass/density = 72.61/(5.32*106) = 13.65*10-6 m3/mol

Number of atoms per unit volume = 6.023*1023/13.65*10-6 = 0.44*1029 atoms/m3

One atom forms 4 covalent bonds, so number of covalent bonds present per unit volume = 4*(0.44*1029) = 1.76*1029 bonds/m3

Intrinsic concentration of electrons and holes = ni = 1*1019 m-3

Each bond gives one electron, so number of bonds broken = number of ionized atoms = 1*1019 m-3

Number of unionized atoms = (1.76*1029) - (1*1019) = 1.76*1029 (approx)

Number of ionized atoms/number of unionized atoms = (1*1019)/(1.76*1029) = 0.568*10-10 = 5.68*10-9

Number of ionized atoms/number of unionized atoms = 5.68*10-9

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