A potential barrier of 0.4 ,V exists across a | vedclass

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(C) The potential barrier opposes the motion of the electron moving from the $n$-side to the $p$-side. The work done by the electric field on the elec

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(C) The potential barrier opposes the motion of the electron moving from the $n$-side to the $p$-side. The work done by the electric field on the electron is negative.Using the work-energy theorem:$W = \Delta K = K_f - K_i$$-e V = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$Here,$V = 0.4 \,V$,$u = 6.0 	imes 10^5 \,ms^{-1}$,$m = 9 	imes 10^{-31} \,kg$,and $e = 1.6 	imes 10^{-19} \,C$.$- (1.6 	imes 10^{-19}) 	imes 0.4 = \frac{1}{2} 	imes (9 	imes 10^{-31}) 	imes (v^2 - (6.0 	imes 10^5)^2)$$-0.64 	imes 10^{-19} = 4.5 	imes 10^{-31} 	imes (v^2 - 36 	imes 10^{10})$$v^2 - 36 	imes 10^{10} = \frac{-0.64 	imes 10^{-19}}{4.5 	imes 10^{-31}}$$v^2 - 36 	imes 10^{10} = -0.1422 	imes 10^{12} \approx -14.22 	imes 10^{10}$$v^2 = (36 - 14.22) 	imes 10^{10} = 21.78 	imes 10^{10}$$v = \sqrt{21.78} 	imes 10^5 \approx 4.66 	imes 10^5 \,ms^{-1}$.Given $v = \frac{x}{3} 	imes 10^5$,so $\frac{x}{3} \approx 4.66 \implies x \approx 14$.

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