Question

The potential barrier existing across an unbiased $p-n$ junction is $0.2\text{V}$. What minimum kinetic energy a hole should have to diffuse from the $p-$side to the $n-$side if the junction is forward biased with $0.1\text{V}$ ?

• A. $0.2\text{eV}$
• B. $0.1\text{eV}$
• C. $0.3\text{eV}$
• D. $0.4\text{eV}$

Answer 1

The correct option is B $0.1\text{eV}$

In case of unbiased $p-n$ junction, the potential barrier can be shown as

Now, by adding a battery in forward bias, reduces the potential barrier.

Since the battery is $0.1\text{V}$, the potential barrier reduces by $0.1\text{V}$.


As we know that hole is basically absence of electron. So the charge of the hole is,

$q=+e$

Electrostatic potential energy of a hole crossing the potential barrier is given by,

$U=qV=e\left(0.1\text{V}\right)=0.1\text{eV}$

By the conservation of energy, the minimum kinetic energy required will be equal to the electrostatic potential energy,

$K{E}_{min}=U$

$\therefore K{E}_{min}=0.1\text{eV}$

Hence, option $\left(\text{B}\right)$ is correct.

Why this question? Note: In forward bias, the battery reduces the barrier across the junction for flow of current