Question

The maximum wavelength of electromagnetic radiation which can create a hole-electron pair in the semiconductor Ge is: (Given, the band gap of Ge is 0.72 eV)

• A. \(\sim 1.1 \times 10^{-5}\) m
• B. \(\sim 1.4 \times 10^{-5}\) m
• C. \(\sim 1.7 \times 10^{-6}\) m
• D. \(\sim 2.6 \times 10^{-6}\) m

Hint:

Use \(\lambda_{max} = hc/E_g\) with \(hc = 1240\) eV nm.

Answer 1

The Correct Option is C

Step 1: Understand the concept:
A hole-electron pair forms when a photon lifts an electron from the valence band to the conduction band. The photon energy must be at least the band gap \(E_g\). Larger wavelength means lower energy, so the maximum wavelength matches photon energy equal to \(E_g\).

Step 2: Write the formula:
Photon energy is \(E = \dfrac{hc}{\lambda}\). At the limit, \(E = E_g\), so \[ \lambda_{max} = \frac{hc}{E_g} \] A handy value is \(hc = 1240\) eV nm.

Step 3: Put in the numbers:
\[ \lambda_{max} = \frac{1240 \text{ eV nm}}{0.72 \text{ eV}} \approx 1722 \text{ nm} \]
Now convert: \(1722\) nm \(= 1.722 \times 10^{-6}\) m.

Step 4: Match with the options:
The value \(1.72 \times 10^{-6}\) m is close to \(1.7 \times 10^{-6}\) m, which is option (3).
Options (1) and (2) are about ten times too large. Option (4) is too large by about 50 percent, so it would need a smaller band gap of about 0.48 eV.

Final Answer:
The maximum wavelength is about \(1.7 \times 10^{-6}\) m, option (3). \[ \boxed{\sim 1.7 \times 10^{-6}\ \text{m}} \]