When a p - n junction is reverse-biased, the current is nearly...

The total current $I$ in a $p$-$n$ junction diode is given by the difference between the diffusion current and the drift current:

$I = I_{\text{diffusion}} - I_{\text{drift}}$

The drift current is due to minority carriers and remains nearly constant. From the given data, the reverse-biased current is constant at $25 \mu\text{A}$, which gives:

$I_{\text{drift}} = 25 \mu\text{A}$

When the diode is reverse-biased, the potential barrier widens, preventing the diffusion of majority carriers across the junction. Consequently, the total current is almost entirely due to the drift current.

Thus, the total current in reverse bias is $I \approx -25 \mu\text{A}$. Substituting this into the current equation:

$-25 \mu\text{A} = I_{\text{diffusion}} - 25 \mu\text{A}$

$I_{\text{diffusion}} = 0$

The forward-bias data is extra information and is not required for this calculation.

Answer: $0$