A semiconductor $\mathrm{X}$ is made by doping silicon with phosphorous. A second semiconductor $\mathrm{Y}$ is made by doping silicon with aluminium. The two are joined by a suitable technique to form a $\mathrm{p}$-$\mathrm{n}$ junction and is connected to a battery such that $\mathrm{Y}$ is joined to negative of the battery and $\mathrm{X}$ to the positive of the battery. Which of the following statements is correct?

Silicon doped with phosphorous (a pentavalent impurity) forms an n-type semiconductor, denoted as X. Silicon doped with aluminium (a trivalent impurity) forms a p-type semiconductor, denoted as Y.

The junction formed is a p-n junction where Y is the p-side and X is the n-side. The p-side (Y) is connected to the negative terminal of the battery, and the n-side (X) is connected to the positive terminal of the battery. This configuration corresponds to reverse biasing of the p-n junction.

In reverse bias, the potential barrier at the junction increases because the external electric field opposes the diffusion of majority carriers across the junction. Consequently, the width of the depletion region increases.

The current that flows in a reverse-biased p-n junction is extremely small and is primarily due to the drift of minority charge carriers (electrons from the p-side and holes from the n-side) across the junction.

Answer: Potential barrier of the junction is raised and current is due to minority carriers