A potential barrier of 0.4 V exists across a p-n junction. An electron enters the junction from the $$n$$-side with a speed of $$6.0 \times 10^5$$ m s$$^{-1}$$. The speed with which electron enters the $$p$$ side will be $$\frac{x}{3} \times 10^5$$ m s$$^{-1}$$, then the value of $$x$$ is ______.: (Given mass of electron $$= 9 \times 10^{-31}$$ kg, charge on electron $$= 1.6 \times 10^{-19}$$ C.)

We need to find the speed of an electron after crossing a potential barrier at a p-n junction. When an electron crosses a potential barrier of V = 0.4 V from n-side to p-side, it loses kinetic energy equal to eV. By conservation of energy, $$\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 - eV$$.

Substituting the values $$v_i = 6.0 \times 10^5$$ m/s, $$m = 9 \times 10^{-31}$$ kg, $$e = 1.6 \times 10^{-19}$$ C, V = 0.4 V, we have $$\frac{1}{2}mv_i^2 = \frac{1}{2} \times 9 \times 10^{-31} \times (6 \times 10^5)^2 = \frac{1}{2} \times 9 \times 10^{-31} \times 36 \times 10^{10} = 162 \times 10^{-21} = 1.62 \times 10^{-19} \text{ J}$$ and the energy lost is $$eV = 1.6 \times 10^{-19} \times 0.4 = 0.64 \times 10^{-19} \text{ J}$$.

Thus the final kinetic energy is $$\frac{1}{2}mv_f^2 = 1.62 \times 10^{-19} - 0.64 \times 10^{-19} = 0.98 \times 10^{-19} \text{ J}$$, so $$v_f^2 = \frac{2 \times 0.98 \times 10^{-19}}{9 \times 10^{-31}} = \frac{1.96 \times 10^{-19}}{9 \times 10^{-31}} = 0.2178 \times 10^{12} = 2.178 \times 10^{11}$$ and $$v_f = \sqrt{2.178 \times 10^{11}} \approx 4.667 \times 10^5 \text{ m/s} = \frac{14}{3} \times 10^5 \text{ m/s}$$.

So $$x = 14$$. The answer is 14.