P-N Junction Diode and its applications: Practice Problem & Solution

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of P-N Junction Diode and its applications. Understanding the underlying formula is key to arriving at the correct answer below:

The minimum frequency is $\nu = \frac{E_g}{h}$. Converting energy to Joules gives $E_g = 2.0 \times 1.6 \times 10^{-19} J$. Thus $\nu = \frac{3.2 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 4.8 \times 10^{14}$ Hz, which is nearly $5 \times 10^{14}$ Hz.