A potential barriers of 0.5 eV exists in p-n junction. If an electron approaches the p-n junction from the n – side with speed of $5\times {10}^{5}m/s$ ,
then electron will enter the P side with velocity (in m/s) 

Let electron enters the n region with speed $v_1$ and enters the P region with speed  $v_2$

Applying conservation of energy, 

                                                        $\frac{1}{2}m{v_1}^2=eV+\frac{1}{2}m{v_2}^2$

$\Rightarrow \frac{1}{2}\times 9.1\times {10}^{-31}\times {(5\times {10}^5)}^2=1.6\times {10}^{-19}\times 0.5+\frac{1}{2}\times 9.1\times {10}^{-31}{v_2}^2$

$\Rightarrow 113.75\times {10}^{-21}=0.8\times {10}^{-19}+\frac{1}{2}\times 9.1\times {10}^{-31}{v_2}^2$

$\Rightarrow 4.55\times {10}^{-31}{v_2}^2=113.75\times {10}^{-21} - 80\times {10}^{-21}$

$\Rightarrow {v_2}^2=\frac{33.75\times {10}^{-21}}{4.55\times {10}^{-31}}$

$\Rightarrow {v_2}^2 = 7.41\times {10}^{10}$

$\Rightarrow v_2 = 2.7\times {10}^5$

$\Rightarrow v_2=27\times {10}^{4}m/s$