Suppose a pure Si crystal has 5 × 10²8 atoms per m³. It is doped with 5 × 10²2 atoms per m³ of Arsenic. Calculate the majority and minority carrier concentration in the doped silicon. (Given : n_i = 1·5 × 10^16\ m^-3)

Electrons (majority) ≈ 5 × 10²2 m^-3; holes (minority) = 4.5 × 10^9 m^-3

PhysicsCBSE Class 12 · Board exam 2026, set 55/5/1 · 2 marks · Very short answer

CBSE Class 12 Physics 2026: question 21

Set in the CBSE Class 12 board exam of 2026, paper 55/5/1, section B, question 21, for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q21 · 2 marks · Very short answerSemiconductor Electronics: Materials, Devices and Simple Circuits

Suppose a pure Si crystal has $5 \times 1 0^{28}$ atoms per $m^{3}$ . It is doped with $5 \times 1 0^{22}$ atoms per $m^{3}$ of Arsenic. Calculate the majority and minority carrier concentration in the doped silicon. (Given : $n_{i} = 1 \cdot 5 \times 1 0^{16} m^{- 3}$ )

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 55/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Arsenic is a donor (group 15) — what type of semiconductor results?

Hint 2

Use the law of mass action $n_{e} n_{h} = n_{i}^{2}$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Arsenic is pentavalent, so each As atom donates one electron. Since $N_{D} = 5 \times 1 0^{22} ≫ n_{i}$ , majority carriers (electrons): $n_{e} \approx N_{D} = 5 \times 1 0^{22} m^{- 3}$
1 mark
Using $n_{e} n_{h} = n_{i}^{2}$ : $n_{h} = \frac{( 1.5 \times 1 0 ^{16} ) ^{2}}{5 \times 1 0 ^{22}} = \frac{2.25 \times 1 0 ^{32}}{5 \times 1 0 ^{22}} = 4.5 \times 1 0^{9} m^{- 3}$ (minority carriers: holes)
1 mark

Answer: Electrons (majority) ≈ 5 × 10^22 m^-3; holes (minority) = 4.5 × 10^9 m^-3

Where the marks are

Majority electron concentration ≈ $N_{D} = 5 \times 1 0^{22}$ m⁻³ (1 mark)
Uses mass action law to get holes $4.5 \times 1 0^{9}$ m⁻³ (1 mark)

Revise the chapter

Chapter: Semiconductor Electronics: Materials, Devices and Simple Circuits (notes, formulas, where marks are lost and every past question)

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CBSE Class 12 Physics 2026: question 21

Hints

Step-by-step solution

Revise the chapter

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