An electron in the n-region of a p-n junction | vedclass

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Question

(A) The initial kinetic energy of the electron is $K_i = \frac{1}{2}mv^2$. As the electron moves through the barrier potential $V$, it loses energy eq

Vedclass · Accepted Answer

$3 	imes 10^5 \ m/s$

Vedclass · Answer

(A) The initial kinetic energy of the electron is $K_i = \frac{1}{2}mv^2$. As the electron moves through the barrier potential $V$, it loses energy equal to $eV$. The final kinetic energy $K_f$ is given by $K_f = K_i - eV = \frac{1}{2}mv_f^2$. Substituting the values: $K_i = \frac{1}{2} 	imes (9 	imes 10^{-31}) 	imes (5 	imes 10^5)^2 = 0.5 	imes 9 	imes 10^{-31} 	imes 25 	imes 10^{10} = 112.5 	imes 10^{-21} \ J$. The energy lost is $eV = 1.6 	imes 10^{-19} 	imes 0.45 = 0.72 	imes 10^{-19} = 72 	imes 10^{-21} \ J$. So, $K_f = 112.5 	imes 10^{-21} - 72 	imes 10^{-21} = 40.5 	imes 10^{-21} \ J$. Now, $\frac{1}{2}mv_f^2 = 40.5 	imes 10^{-21} \implies v_f^2 = \frac{2 	imes 40.5 	imes 10^{-21}}{9 	imes 10^{-31}} = 9 	imes 10^{10} \ m^2/s^2$. Therefore, $v_f = 3 	imes 10^5 \ m/s$.

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