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

# Resistance of antimony-doped germanium bar

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2074 Asoj, 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-antimony-bar
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Four micrograms of antimony are thoroughly mixed in molten form with 100 gms of pure germanium. Find the density of antimony atoms, density of donated electrons and the total resistance of a bar of such n-type material of 2 cm long, 0.012×0.012 cm in cross-section. Take, density of Ge = 5.46 gm/cm³ and atomic weight of Sb = 121.76.

## Short answer

Sb atom density in Ge is N_d = (4×10^-6 g / 121.76 g/mol × N_A) / (100 g / 5.46 g/cm^3) = 1.08×10^15 cm^-3, all ionized as electrons. With μ_e=3900 cm^2/V·s, resistance R = ρL/A = 20.6 kΩ.

## Answer

Antimony (group V) is a donor; each Sb atom donates one electron.

**Given:** $m_{Ge} = 100$ g, $\rho_{Ge} = 5.46\ \text{g cm}^{-3}$, $m_{Sb} = 4\ \mu\text{g} = 4\times10^{-6}$ g, $M_{Sb} = 121.76$; bar L = 2 cm, $A = 0.012\times0.012 = 1.44\times10^{-4}\ \text{cm}^2$. **Assumed:** electron mobility in Ge $\mu_e = 3900\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$ (standard value, not given), $N_A = 6.022\times10^{23}$.

#### Step 1: Volume of germanium

$$
V = \frac{100}{5.46} = 18.32\ \text{cm}^3
$$

#### Step 2: Number of Sb atoms

$$
N_{Sb} = \frac{4\times10^{-6}}{121.76}\times6.022\times10^{23} = 1.978\times10^{16}
$$

#### Step 3: Density of antimony atoms

$$
N_d = \frac{1.978\times10^{16}}{18.32} = 1.08\times10^{15}\ \text{cm}^{-3}
$$

#### Step 4: Density of donated electrons

At room temperature every Sb atom is ionised, and $N_d \gg n_i(\text{Ge}) = 2.4\times10^{13}$, so

$$
n \approx N_d = 1.08\times10^{15}\ \text{cm}^{-3}
$$

#### Step 5: Conductivity and resistance

$$
\begin{aligned}
\sigma &= eN_d\mu_e = 1.6\times10^{-19}\times1.08\times10^{15}\times3900\\
&= 0.674\ \text{S cm}^{-1}\\
\rho &= 1/\sigma = 1.484\ \Omega\,\text{cm}\\
R &= \frac{\rho L}{A} = \frac{1.484\times2}{1.44\times10^{-4}}\\
&= 2.06\times10^{4}\ \Omega
\end{aligned}
$$

| Quantity | Value |
|---|---|
| Sb atom density | $1.08\times10^{15}\ \text{cm}^{-3}$ |
| Donated electrons | $1.08\times10^{15}\ \text{cm}^{-3}$ |
| Resistivity | 1.48 Ω cm |
| Resistance | 20.6 kΩ |

**Answer: $N_d = n \approx 1.08\times10^{15}\ \text{cm}^{-3}$; $R \approx 20.6\ \text{k}\Omega$** (with $\mu_e = 3800$ the resistance would be about 21.1 kΩ).

## 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 obtained?
   A. Mass divided by density: 100/5.46
   B. Mass times density
   C. Density divided by mass
   D. Number of atoms divided by N_A
   Answer: A. V = m/rho = 100/5.46 = 18.32 cm^3. (Key point: Volume of Ge)

2. How is the number of Sb atoms found from 4 micrograms?
   A. (M / m) x N_A
   B. m x M x N_A
   C. (m / M) x N_A
   D. m x N_A / density
   Answer: C. Atoms = (mass/atomic weight) x Avogadro number = 1.98 x 10^16. (Key point: Number of Sb atoms)

3. What is the density of antimony atoms (and of donated electrons)?
   A. 1.98 x 10^16 cm^-3
   B. 1.08 x 10^15 cm^-3
   C. 1.08 x 10^15 cm^-3 for atoms but 2.4 x 10^13 for electrons
   D. 5.46 x 10^15 cm^-3
   Answer: B. N_d = 1.978 x 10^16 / 18.32 = 1.08 x 10^15 cm^-3, and since N_d >> n_i each atom gives one electron. (Key point: Donor density)

4. Why can the electron density be taken equal to N_d?
   A. Because each Sb is ionised at room temperature and N_d >> n_i of Ge
   B. Because Sb is an acceptor
   C. Because germanium has no intrinsic carriers
   D. Because mobility is high
   Answer: A. All donors ionise and they far outnumber thermally generated carriers. (Key point: Why n = N_d)

5. Which sequence gives the resistance?
   A. R = sigma L / A, sigma = e N_d mu_e
   B. R = rho A / L, rho = 1/sigma
   C. R = rho L / A, rho = 1/(e N_d mu_e)
   D. R = e N_d mu_e L / A
   Answer: C. R = rho L/A with rho = 1/(e N_d mu_e) = 1.48 ohm cm, giving about 20.6 kilo-ohm. (Key point: Resistance of the bar)

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