Verified Solution • 2025–26 Session moderate

QPages 419-421: Q8: HC Verma: Concepts of Physics - Volume 2 Semiconductors and Semiconductor Devices Pages 419-421: Q8 Solution: Find the maximum wavelength of electromagnetic radiation which…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often write $\lambda = h/E$ and insert $h$ in eV·s, forgetting the speed‑of‑light factor; the correct formula is $\lambda = hc/E$, so you must either use $c=3\times10^8\,$m/s with $h$ in joule·s or the combined constant $hc=1240\,$eV·nm. Omitting $c$ gives a wavelength off by about $10^8$ and leads to nonsensical answers such as 19 m instead of $\sim1.9\times10^{-6}$ m.

Find the maximum wavelength of electromagnetic radiation which can create a hole-electron pair in germanium. The band gap in germanium is 0.65 eV. Use Planck constant h=4.14×10−15 eV-sh = 4.14 \times 10^{-15} \text{ eV-s}, Boltzmann constant k=8.62×10−5 eV/Kk = 8.62 \times 10^{-5} \text{ eV/K}.)

Step-by-Step Verified Solution

**Answer:** Maximum wavelength = 1.9 _ 10__ m = 19 _m. **Step-by-Step Explanation:** Given: Band gap = 0.65 eV.
λmax=12400.65=1907 nm=1.9×10−6 m\lambda_{max} = \frac{1240}{0.65} = 1907 \text{ nm} = 1.9 \times 10^{-6} \text{ m}
Standard answer: 1.9 _ 10__ m (using different conversion).

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