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: Physical picture:
To free a bound electron in Ge, the incoming photon must give at least 0.72 eV. Photons of longer wavelength than a limiting value carry less energy than that, so they cannot make a pair.

Step 2: Use SI units:
Take $h = 6.63 \times 10^{-34}$ J s, $c = 3 \times 10^{8}$ m/s and $1 \text{ eV} = 1.6 \times 10^{-19}$ J.
Band gap $E_g = 0.72 \times 1.6 \times 10^{-19} = 1.152 \times 10^{-19}$ J.

Step 3: Calculate:
\[ \lambda = \frac{hc}{E_g} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1.152 \times 10^{-19}} \]
The numerator is $1.989 \times 10^{-25}$. So \[ \lambda = 1.727 \times 10^{-6} \text{ m} \]

Step 4: Pick the option:
This is closest to $1.7 \times 10^{-6}$ m, which is the third option. The others differ by a large factor.

Final Answer:
The answer is option (3). \[\boxed{\lambda_{max} \approx 1.7 \times 10^{-6}\ \text{m}}\]