Consider a n-type semiconductor in which ne and nh are number of electrons and holes, respectively. (A) Holes are minority carriers (B) The dopant is a pentavalent atom (C) ne nh != ni^2 (where ni is number of electrons or holes in semiconductor when it is intrinsic form) (D) ne nh >= ni^2 (E) The holes are not generated due to the donors Choose the correct answer from the options given below :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q43 – JEE-MAIN 2025

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Home Physics SemiconductorsQuestion #471

Consider a n-type semiconductor in which

n_{e}

and

n_{h}

are number of electrons and holes, respectively. (A) Holes are minority carriers (B) The dopant is a pentavalent atom (C)

n_{e} n_{h} \neq n_{i}^{2}

(where

n_{i}

is number of electrons or holes in semiconductor when it is intrinsic form) (D)

n_{e} n_{h} \geq n_{i}^{2}

(E) The holes are not generated due to the donors Choose the correct answer from the options given below :

Clue 1 of 3: Understand n-type semiconductors

Recall the fundamental properties of n-type extrinsic semiconductors, including majority/minority carriers and the nature of dopants.

🥷

Ninja Strategy•Mass Action Law Check

Quickly recall the Mass Action Law ( $n_{e} n_{h} = n_{i}^{2}$ ) to eliminate any options that state inequalities or non-equality for this product, leaving only the correct combination.

🧠 Full Solution Path

Step 1: Analyze the properties of an n-type semiconductor✦ Active

In an n-type semiconductor, pentavalent impurities (donors) are added to an intrinsic semiconductor. These donors contribute extra electrons to the conduction band, making electrons the majority carriers and holes the minority carriers. Holes are primarily generated due to thermal excitation, not directly by the donor atoms.

Based on this: (A) Holes are minority carriers - TRUE (B) The dopant is a pentavalent atom - TRUE (E) The holes are not generated due to the donors - TRUE

Step 2: Apply the Mass Action Law for semiconductors○ Expand

The Mass Action Law states that for any semiconductor in thermal equilibrium, the product of electron concentration ( $n_{e}$ ) and hole concentration ( $n_{h}$ ) is equal to the square of the intrinsic carrier concentration ( $n_{i}$ ). This law holds true for both intrinsic and extrinsic semiconductors.

n_{e} n_{h} = n_{i}^{2}

Based on this: (C) $n_{e} n_{h} \neq n_{i}^{2}$ - FALSE (D) $n_{e} n_{h} \geq n_{i}^{2}$ - FALSE (it must be exactly equal, not greater than or equal to, unless there's a non-equilibrium condition, which is not implied here).

💡 Teacher's Secret Hint

Ensure you recall the exact form of the Mass Action Law. It's an equality, not an inequality.

Step 3: Identify the correct combination of statements○ Expand

From the analysis, statements (A), (B), and (E) are true. Statements (C) and (D) are false. Therefore, the correct option is the one that includes (A), (B), and (E).

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 43 – JEE-MAIN 2025

Clue 1 of 3: Understand n-type semiconductors

🧠 Full Solution Path