i = 2/100 = 20 mA 

In Reverse bias mode potential of p-type is less than that of n-type semiconductor

For Forward biasing of PN Junction diode potential of P side should be greater than potential of N-side semiconductor

Energy of EM wave E = hυ

⇒ 2 eV = 6.626 × 10^-34×υ

⇒  \[v= 5\times 10_{14} Hz\]

Factual Question: When a PN junction diode is reverse biased potential barrier of depletion increases. Thus Electrons and holes move away from the junction depletion region.

Theory: In forward biased mode diffusion current is more than drift current. Where as in reverse biased mode diffusion current decreased and become less than drift current.

Diode D2 and D3 are forward biased. Where as Diode D1 is reversed biased. No current will flow through D1. 

Equivalent Resistance= 20/3 ohm

Current = Voltage / Resistance = 10/ (20/3) = 1.5 Ampere

A Photodiode and Zener Diode operate in Reverse Bias, while a Solar Cell operates in an Unbiased condition (zero external voltage).

• Photodiode: Works in reverse bias to detect light.
• Zener Diode: Operates in reverse bias for voltage regulation.
• Solar Cell: Generates power in an unbiased condition when exposed to light.

In forward bias, the applied field opposes the barrier field, pushing major carriers (electrons on n-side and holes on p-side) towards the junction, thereby decreasing the depletion width. Thus, statement (4) is incorrect.

Forward resistance of a diode is very small (\(approx 10\Omega\)), while its reverse resistance is extremely high (\(approx 10^5 \Omega\)). The ratio \(R_f / R_r\) is of the order of \(10^{-4}\), which corresponds to \(1 : 10^4\).