A P-N junction has a potential barrier of 0.5 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The potential barrier $V = 0.5 \ V$. The mass of an electron $m = 9.1 	imes 10^{-31} \ kg$ and its charge $e = 1.6 	imes 10^{-19} \ C$.According

Vedclass · Accepted Answer

$2.7 	imes 10^5 \ m/s$

Vedclass · Answer

(D) The potential barrier $V = 0.5 \ V$. The mass of an electron $m = 9.1 	imes 10^{-31} \ kg$ and its charge $e = 1.6 	imes 10^{-19} \ C$.According to the law of conservation of energy,the initial kinetic energy of the electron is equal to the final kinetic energy plus the potential energy barrier it must overcome:$\frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2 + eV$$\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 - eV$$v_f^2 = v_i^2 - \frac{2eV}{m}$$v_f = \sqrt{v_i^2 - \frac{2eV}{m}}$Substituting the values:$v_f = \sqrt{(5 	imes 10^5)^2 - \frac{2 	imes 1.6 	imes 10^{-19} 	imes 0.5}{9.1 	imes 10^{-31}}}$$v_f = \sqrt{25 	imes 10^{10} - \frac{1.6 	imes 10^{-19}}{9.1 	imes 10^{-31}}}$$v_f = \sqrt{25 	imes 10^{10} - 0.1758 	imes 10^{12}}$$v_f = \sqrt{25 	imes 10^{10} - 17.58 	imes 10^{10}}$$v_f = \sqrt{7.42 	imes 10^{10}} \approx 2.72 	imes 10^5 \ m/s$.

$A$ $P-N$ junction has a potential barrier of $0.5 \ V$. If an electron enters the $P-N$ junction diode from the $N$-side with a speed of $5 \times 10^5 \ m/s$,what will be its speed when it emerges from the $P$-side?

Similar Questions

How does the height of the potential barrier change in forward bias and reverse bias?

$A$ potential barrier of $0.50 \ V$ exists across a $P-N$ junction. If the depletion region is $5.0 \times 10^{-7} \ m$ wide,the intensity of the electric field in this region is:

The barrier potential of a $P-N$ junction diode does not depend on:

Application of a forward bias to a $P-N$ junction

In the given circuit, the current through the battery is ...... $A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module