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

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(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

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$2.7 	imes 10^5 \ m/s$

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(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?

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