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A silicon crystal is doped with Boron atoms. A small electric field points from left to right On an average, the Boron ions i
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The atomic number of Boron is 5 and it is one of the element in the 13th group. So when Boron is doped into the Silicon crystal, the crystal will be changed into P-type semiconductor as shown in the diagram below.

Si OO Si OO Si Silicon (Si): Four valence electrons Si Adding boron to pure silicon crystal results in lack of an electron. A

If a small amount of boron is doped to a single crystal of silicon, valence electrons will be insufficient at one position to bond silicon and boron, resulting in holes* that lack electrons. When a voltage is applied in this state, the neighboring electrons move to the hole, so that the place where an electron is present becomes a new hole, and the holes appear to move to the "–" electrode in sequence. That is how a P-type material is formed.

Holes p-type (B+ O Electron +Now an electric field is applied across the P-type semi conductor formed by Boron deposition as shown in the diagram below, ie, the electric field points from left to right as told in the question. The Boron atom that lost the electron will gain a positive charge and becomes Boron ions (B+ ). Since these ions are positively charged particles, they will move towards the negative electrode of the electric field. Which means the Boron Ions formed will follow the electric field's same direction and move from left to right. Even with out any electric field, the moment of the positive B+ ion will be directed in the same direction of the electric field of the P-type semi conductor. Since the majority carriers are holes in the P-type materials, the positive Boron ions too will follow this holes moment.

So the answer is option (a). The Boron ions will move from left to right in the direction of electric field

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