Consider a p-n junction diode when it is forward biased with 2 V. Which of the following is/are the correct magnitude(s) of the energy difference between quasi Fermi-levels, E_fn in the n-side and E_fp in the p-side?

The correct option is 2 eV.

To understand why this is the correct answer, let us analyze the behavior of the quasi-Fermi levels in a forward-biased p-n junction diode.

When a p-n junction is in thermal equilibrium (no external applied voltage), the Fermi levels on the n-side ($E_{\mathrm{Fn}}$) and the p-side ($E_{\mathrm{Fp}}$) are aligned, meaning there is a single, constant Fermi energy level across the entire device:
$E_{\mathrm{Fn}}-E_{\mathrm{Fp}}=0$

When an external forward bias voltage $V$ is applied across the p-n junction, the system is no longer in thermal equilibrium. Under these non-equilibrium conditions, the Fermi level splits into two quasi-Fermi levels:
1. $E_{\mathrm{fn}}$ which describes the carrier concentration of electrons in the conduction band.
2. $E_{\mathrm{fp}}$ which describes the carrier concentration of holes in the valence band.

The applied potential energy difference shifts the energy bands. Specifically, a forward bias of voltage $V$ lowers the potential barrier of the junction by an energy equivalent to $\mathrm{qV}$, where $q$ is the elementary charge of an electron. This causes a separation between the quasi-Fermi levels in the neutral regions of the n-side and p-side.

The magnitude of the energy difference between the electron quasi-Fermi level on the n-side ($E_{\mathrm{fn}}$) and the hole quasi-Fermi level on the p-side ($E_{\mathrm{fp}}$) is directly proportional to the applied bias voltage $V$ and is given by the relation:
$E_{\mathrm{fn}}-E_{\mathrm{fp}}=qV$

Given in the problem, the forward bias voltage is:
$V=2\text{V}$

Substituting this value into our relation, we obtain the energy difference:
$E_{\mathrm{fn}}-E_{\mathrm{fp}}=q\cdot \left(2\text{V}\right)=2\text{eV}$

Thus, the magnitude of the energy difference between the quasi-Fermi levels is 2 eV.