---
title: "Resistance of arsenic-doped germanium wire"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:germanium-arsenic-wire"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/germanium-arsenic-wire
site: BCT Engineering
---

# Resistance of arsenic-doped germanium wire

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2078 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/germanium-arsenic-wire
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

In a pure germanium of 50g, 5μg of Arsenic is thoroughly mixed in molten form. The density of germanium is 5.46 gcm⁻³ and atomic weight of Arsenic is 74.92 gmol⁻¹. Find the total resistance of a wire of such n-type material having length of 1cm and cross-sectional area of 2.25 × 10⁻⁴cm². Take mobility of electrons in germanium = 3600 cm²V⁻¹s⁻¹.

## Short answer

Donor concentration from As mass and Ge volume is 4.39×10^15 cm⁻³; conductivity σ = e N_d μ_e = 2.53 S/cm, resistivity 0.396 Ω·cm. Wire resistance R = ρL/A ≈ 1.76 kΩ.

## Answer

Arsenic is a donor; each As atom gives one free electron. Find the donor concentration from the masses, then the conductivity and resistance.

**Given:** mass of Ge = 50 g, density 5.46 g cm⁻³; mass of As = 5 µg = $5\times10^{-6}$ g, M = 74.92 g mol⁻¹; $\mu_e = 3600\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$; L = 1 cm; $A = 2.25\times10^{-4}\ \text{cm}^2$; $N_A = 6.022\times10^{23}$.

#### Step 1: Volume of germanium

$$
V = \frac{m}{\rho} = \frac{50}{5.46} = 9.158\ \text{cm}^3
$$

(The tiny mass of As does not change the volume.)

#### Step 2: Number of As atoms

$$
N_{As} = \frac{5\times10^{-6}}{74.92}\times6.022\times10^{23} = 4.019\times10^{16}
$$

#### Step 3: Donor (electron) concentration

$$
N_d = \frac{4.019\times10^{16}}{9.158} = 4.389\times10^{15}\ \text{cm}^{-3}
$$

All donors are ionised at room temperature and $N_d \gg n_i(\text{Ge}) \approx 2.4\times10^{13}$, so $n \approx N_d$ and the hole contribution is negligible.

#### Step 4: Conductivity and resistivity

$$
\begin{aligned}
\sigma &= eN_d\mu_e = 1.6\times10^{-19}\times4.389\times10^{15}\times3600\\
&= 2.528\ \text{S cm}^{-1}\\
\rho &= 1/\sigma = 0.3956\ \Omega\,\text{cm}
\end{aligned}
$$

#### Step 5: Resistance of the wire

$$
\begin{aligned}
R &= \frac{\rho L}{A} = \frac{0.3956\times1}{2.25\times10^{-4}}\\
&= 1758\ \Omega
\end{aligned}
$$

**Answer: $N_d \approx 4.39\times10^{15}\ \text{cm}^{-3}$, $\rho \approx 0.396\ \Omega$ cm, and the wire resistance $R \approx 1.76\ \text{k}\Omega$.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the volume of the germanium found?
   A. V = mass/density = 50/5.46 = 9.158 cm^3
   B. V = 50 x 5.46
   C. V = 5.46/50
   D. V = 50 g / 74.92
   Answer: A. The As mass is negligible so V is the Ge volume. (Key point: Volume)

2. How many As atoms are there?
   A. 5x10^-6 x 6.022x10^23 = 3x10^18
   B. (5x10^-6 / 74.92) x 6.022x10^23 = 4.02x10^16
   C. (74.92/5x10^-6) x 6.022x10^23
   D. 4.02x10^10
   Answer: B. Moles = mass / molar mass; multiply by Avogadro's number. (Key point: Atoms of As)

3. What is Nd and why is n about Nd?
   A. 4.02x10^16 cm^-3, as volume is ignored
   B. 4.39x10^15 cm^-3, and holes dominate
   C. 4.39x10^15 cm^-3, since each As gives one electron and Nd >> ni of Ge
   D. 9.158 cm^-3
   Answer: C. Nd = 4.019x10^16 / 9.158; ionised donors far exceed ni. (Key point: Donor density)

4. How is R found?
   A. R = rho A / L = 0.396 x 2.25e-4
   B. R = rho = 0.396 ohm
   C. R = 1/(e Nd) = 1.4 ohm
   D. sigma = e Nd mu_e = 2.53 S/cm, rho = 0.396 ohm cm, R = rho L/A about 1.76 kohm
   Answer: D. R = 0.3956 x 1 / 2.25x10^-4 = 1758 ohm. (Key point: Wire resistance)

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