1. Use energy conservation across the junction

When an electron moves from the n-side to the p-side across a potential barrier of $0.4\,\text{V}$, its electric potential energy changes by

$\Delta U = q\Delta V = (-e)(+0.4) = -0.4e$

Since the electron has negative charge, moving to higher potential decreases its potential energy, so its kinetic energy increases by

$\Delta K = 0.4e = 0.4 \times 1.6 \times 10^{-19} = 6.4 \times 10^{-20}\,\text{J}$

2. Initial kinetic energy

Given initial speed:

$u = 6.0 \times 10^5\,\text{m s}^{-1}$

So,

$K_i = \frac12 m u^2 = \frac12 \times 9\times 10^{-31} \times (6\times 10^5)^2$

$(6\times 10^5)^2 = 36\times 10^{10} = 3.6\times 10^{11}$

Hence,

$K_i = \frac12 \times 9\times 10^{-31} \times 3.6\times 10^{11}$

$K_i = 16.2\times 10^{-20} = 1.62\times 10^{-19}\,\text{J}$

3. Final kinetic energy

$K_f = K_i + \Delta K$

$K_f = 1.62\times 10^{-19} + 6.4\times 10^{-20}$

$K_f = 2.26\times 10^{-19}\,\text{J}$

4. Find final speed

$\frac12 m v^2 = 2.26\times 10^{-19}$

$v^2 = \frac{2\times 2.26\times 10^{-19}}{9\times 10^{-31}}$

$v^2 = \frac{4.52}{9} \times 10^{12}$

$v^2 \approx 0.5022\times 10^{12} = 5.022\times 10^{11}$

$v \approx \sqrt{5.022\times 10^{11}} \approx 7.09\times 10^5\,\text{m s}^{-1}$

5. Match with the given form

Given

$v = \frac{x}{3}\times 10^5\,\text{m s}^{-1}$

So,

$\frac{x}{3}\times 10^5 = 7.09\times 10^5$

$\frac{x}{3} = 7.09$

$x \approx 21.27$

Since the answer is expected as an integer, we take

$x = 21$

6. Check using a quicker ratio method

Since kinetic energy is proportional to $v^2$,

$\frac12 m v^2 = \frac12 m u^2 + eV$

$v^2 = u^2 + \frac{2eV}{m}$

Substituting:

$v^2 = (6\times10^5)^2 + \frac{2\times1.6\times10^{-19}\times0.4}{9\times10^{-31}}$

$v^2 = 3.6\times10^{11} + 1.422\times10^{11} = 5.022\times10^{11}$

which again gives

$v \approx 7.09\times10^5\,\text{m s}^{-1}$

So the derived integer is 21.

7. Comparison with stored answer

Stored correct answer is 14, but the physically correct calculation gives 21. The stored answer appears inconsistent with energy conservation for an electron crossing a $0.4\,\text{V}$ barrier from n-side to p-side.