In a semiconductor p-n diode, the doping concentrations on p-side...

In a p-n junction diode, the total charge on both sides of the depletion region must be equal to maintain charge neutrality.

Let $x_p$ and $x_n$ be the widths of the depletion region on the p-side and n-side, respectively. Let $N_A$ and $N_D$ be the doping concentrations on the p-side and n-side.

The charge neutrality condition is given by:
$N_A x_p = N_D x_n$

Given $N_A = 10^{15} \text{ atoms/cm}^3$ and $N_D = 10^{18} \text{ atoms/cm}^3$.

Substituting the values, we get:
$10^{15} x_p = 10^{18} x_n$

$x_p = 10^3 x_n$

Since $x_p > x_n$, the depletion region penetrates more into the lightly doped side. Therefore, the depletion region width is more on the p-side compared to that in the n-side.

Answer: The depletion region width is more on p-side compared to that in n-side