Question

A potential barrier of $0.50\text{V}$ exists across a $p-n$ junction. The depletion region is $5\times {10}^{-7}\text{m}$ wide. An electron with speed $5\times {10}^5{\text{ms}}^{-1}$ approaches the $p-n$ junction from the $n-$ side with what speed will it enter the $p-$ side ?

• A. $1.5\times {10}^5{\text{ms}}^{-1}$
• B. $2.7\times {10}^5{\text{ms}}^{-1}$
• C. $3.5\times {10}^5{\text{ms}}^{-1}$
• D. $4.7\times {10}^5{\text{ms}}^{-1}$

Answer 1

The correct option is B $2.7\times {10}^5{\text{ms}}^{-1}$

Given that,
Potential barrier, $V=0.50\text{V}$
width of deflection region, $d=5\times {10}^{-7}\text{m}$
Initial speed of electron, $v_1=5\times {10}^5{\text{ms}}^{-1}$

Let the velocity of electron is $v_2$ when it comes out of depletion region.

The electric field in the $p-n$ junction exists in the direction $n\to p$.

So, an electron crossing the $p-n$ junction will experience an opposing force due to electric field.

Therefore, there will be increase in the electrostatic potential energy of the electron.

Increase in Potential energy $U=e\times V$

$\Rightarrow U=e\times 0.50\text{V}=0.5\text{eV}$

$\Rightarrow U=0.5\times 1.6\times {10}^{-19}\text{J}$

$\Rightarrow U=0.8\times {10}^{-19}\text{J}$

By using principle of conservation of energy

$\frac{1}{2}m{v}_1^2=U+\frac{1}{2}m{v}_2^2$

$\Rightarrow \frac{1}{2}\times (9.1\times {10}^{-31})\times (5\times {10}^5{)}^2=0.8\times {10}^{-19}+\frac{1}{2}(9.1\times {10}^{-31})v_2^2$

$\Rightarrow 1.13\times {10}^{-19}=0.8\times {10}^{-19}+(4.55\times {10}^{-31})v_2^2$

$\Rightarrow v_2^2=\frac{0.33\times {10}^{-19}}{4.55\times {10}^{-31}}$

$\therefore v_2=2.7\times {10}^5{\text{ms}}^{-1}$

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

Why this question? Tip:The direction of electric field in the depletion region of $p-n$ junction is from $n$ to $p$ side.