Physics - Semiconductors Question with Solution | TestHub

Applying conservation of energy

$-e{V}_n+\frac{1}{2}m{v}_i^2=-e{V}_p+\frac{1}{2}m{v}_f^2$

We know, $V_n-V_p=$potential barrier$=0.4 \mathrm{V}$

Then, we have, $-e\left(V_n-V_p\right)+\frac{1}{2}m{v}_i^2=\frac{1}{2}m{v}_f^2$

$\Rightarrow -1.6\times {10}^{-19}\times 0.4+\frac{1}{2}\times 9\times {10}^{-31}\times 36\times {10}^{10}=\frac{1}{2}\times 9\times {10}^{-31}{ \mathrm{v}}_f^2$

$\Rightarrow -0.64\times {10}^{-19}+1.62\times {10}^{-19}=\frac{1}{2}\times 9\times {10}^{-31}{v}_f^2$

$\Rightarrow 0.98\times {10}^{-19}=4.5\times {10}^{-31}{v}_f^2$

$\Rightarrow v_f^2=\frac{0.98\times {10}^{12}}{4.5}$

$\Rightarrow v_f=0.466666\times {10}^6$

$=4.6666\times {10}^5 \mathrm{m} {\mathrm{s}}^{-1}=\frac{14}{3}\times {10}^5 \mathrm{m} {\mathrm{s}}^{-1}$