Question 57 of 209

A potential barrier of $0.4V$ exists across a p-n junction. An electron enters the junction from the $n$-side with a speed of $6.0\times {10}^{5}{ms}^{-1}$. The speed with which electron enters the $p$ side will be $\dfrac{x}{3}\times {10}^{5}{ms}^{-1}$, then the value of $x$ is _____.: (Given mass of electron $=9\times {10}^{-31}kg$, charge on electron $=1.6\times {10}^{-19}C$.)

Your answerCheck answer

Show answer and solution▾

Answer: 14

Applying conservation of energy
$-e{V}_{n}+\dfrac{1}{2}m{v}_{i}^{2}=-e{V}_{p}+\dfrac{1}{2}m{v}_{f}^{2}$
We know, ${V}_{n}-{V}_{p}=$potential barrier$=0.4V$
Then, we have, $-e\left({V}_{n}-{V}_{p}\right)+\dfrac{1}{2}m{v}_{i}^{2}=\dfrac{1}{2}m{v}_{f}^{2}$
$\Rightarrow -1.6\times {10}^{-19}\times 0.4+\dfrac{1}{2}\times 9\times {10}^{-31}\times 36\times {10}^{10}=\dfrac{1}{2}\times 9\times {10}^{-31}{v}_{f}^{2}$
$\Rightarrow -0.64\times {10}^{-19}+1.62\times {10}^{-19}=\dfrac{1}{2}\times 9\times {10}^{-31}{v}_{f}^{2}$
$\Rightarrow 0.98\times {10}^{-19}=4.5\times {10}^{-31}{v}_{f}^{2}$
$\Rightarrow {v}_{f}^{2}=\dfrac{0.98\times {10}^{12}}{4.5}$
$\Rightarrow {v}_{f}=0.466666\times {10}^{6}$
$=4.6666\times {10}^{5}m{s}^{-1}=\dfrac{14}{3}\times {10}^{5}m{s}^{-1}$