Question

How can p-type and n-type crystal from pure semiconductor be formed? Find the maximum wavelength of electromagnetic radiation which can create a hole-electron pair in germanium of a band-gap 0.65 eV.
OR
How is the conductivity of a metal and semiconductor changed with raising temperature? Calculate the conductivity of an n-type semiconductor from the following data: Density of conduction electrons = 8×10¹³ /cm³, Density of holes = 5×10¹² /cm³, Mobility of conduction electrons = 2.3×10⁴ cm²/V·s, Mobility of holes = 100 cm²/V·s.

Hint:

Doping with a pentavalent impurity gives n-type (extra electrons) and a trivalent impurity gives p-type (holes). Use \( \lambda_{max} = hc/E_g \) for the germanium part, and \( \sigma = e(n_e\mu_e + n_h\mu_h) \) for the conductivity.

Answer 1

Option 1: Doping seen through energy bands, and the threshold wavelength

Step 1: A pure semiconductor at 0 K has a completely filled valence band and an empty conduction band separated by a gap \( E_g \). Doping inserts new allowed levels inside this gap. A pentavalent (donor) atom places filled donor levels just below the conduction band; their electrons jump up very easily, so conduction is mainly by electrons and the crystal is n-type.

Step 2: A trivalent (acceptor) atom places empty acceptor levels just above the valence band; valence electrons jump into them and leave holes behind, so conduction is mainly by holes and the crystal is p-type. In both cases only a tiny impurity fraction (about 1 atom in 10⁶) changes the conductivity enormously.

Step 3: A photon can generate an electron-hole pair only if its energy is not less than the band-gap. The limiting (maximum-wavelength) photon has energy exactly \( E_g \). Using the handy relation with energy in eV and wavelength in micrometre, \[ \lambda_{max}(\mu\text{m}) = \frac{1.24}{E_g(\text{eV})} \]

Step 4: Substitute \( E_g = 0.65\ \text{eV} \): \[ \lambda_{max} = \frac{1.24}{0.65} = 1.9\ \mu\text{m} \]

Step 5: Cross-check from first principles: \[ \lambda_{max} = \frac{hc}{E_g} = \frac{1.98\times10^{-25}}{0.65\times1.6\times10^{-19}} = 1.9\times10^{-6}\ \text{m} \] Both routes agree, and the wavelength lies in the infrared.
\[\boxed{\lambda_{max} \approx 1.9\ \mu\text{m} = 1.9\times10^{-6}\ \text{m}}\]

Option 2: Conductivity from separate carrier channels, with a resistivity check

Step 1: The temperature behaviour has opposite sign in the two materials. In a metal the carrier number is essentially fixed while thermal vibrations scatter the electrons more strongly at higher T, so \( \sigma \) falls (resistance grows with temperature). In a semiconductor, extra bonds break as T rises and the carrier number soars, so \( \sigma \) climbs (resistance falls with temperature).

Step 2: The total conductivity is the sum of the electron and hole channels: \[ \sigma = e\,n_e\,\mu_e + e\,n_h\,\mu_h \] Evaluate each channel on its own.

Step 3 (electron channel): \[ \sigma_e = (1.6\times10^{-19})(8\times10^{13})(2.3\times10^{4}) = 0.2944\ \Omega^{-1}\text{cm}^{-1} \]
Step 4 (hole channel): \[ \sigma_h = (1.6\times10^{-19})(5\times10^{12})(100) = 8.0\times10^{-5}\ \Omega^{-1}\text{cm}^{-1} \]
Step 5 (add): \[ \sigma = 0.2944 + 0.00008 = 0.2945 \approx 0.294\ \Omega^{-1}\text{cm}^{-1} \] The hole channel is negligible, which confirms an n-type sample.
Step 6 (SI value and resistivity): \( \sigma = 0.294\ \Omega^{-1}\text{cm}^{-1} \times 100 = 29.4\ \Omega^{-1}\text{m}^{-1} \), so the resistivity is \( \rho = 1/\sigma \approx 3.4\ \Omega\,\text{cm} \) (about 0.034 Ω·m).
\[\boxed{\sigma \approx 0.294\ \Omega^{-1}\text{cm}^{-1} \approx 29.4\ \text{S/m}}\]