---
title: "Lattice constant of a simple cubic structure"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:simple-cubic-lattice-constant"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/simple-cubic-lattice-constant
site: BCT Engineering
---

# Lattice constant of a simple cubic structure

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2074 Asoj, 4 marks
- Video: [Unit Cell Chemistry   Simple Cubic, Body Centered Cubic, Face Centered Cubic Crystal Lattice Structu](https://www.youtube.com/watch?v=HCWwRh5CXYU) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/simple-cubic-lattice-constant
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

Prove that for a simple cubic structure, the lattice constant: a = [NM/(ρNA)]^(1/3) where, N is the number of atoms per unit cell, M is atomic weight, NA is Avogadro's number and ρ is density of crystal material.

## Short answer

Mass per atom = M/N A, mass of unit cell with N atoms = N M/N A. Density ρ = (N M)/(N A a³). Rearranging yields a³ = N M/(ρ N A), hence a = [N M/(ρ N A)]^(1/3). Applies for any cubic lattice using the correct N.

## Answer

The density of a crystal equals the mass of the atoms in one unit cell divided by the volume of the cell.

Let
- $N$ = number of atoms per unit cell,
- $M$ = atomic weight (mass of one mole of atoms, kg/mol),
- $N_A$ = Avogadro's number,
- $a$ = lattice constant, $\rho$ = density.

#### Proof
Mass of one atom:

$$
m_{atom} = \frac{M}{N_A}
$$

Mass of atoms in one unit cell:

$$
m_{cell} = N\cdot\frac{M}{N_A}
$$

The unit cell is a cube of side $a$, so its volume is $V_{cell} = a^3$.

Density:

$$
\rho = \frac{m_{cell}}{V_{cell}} = \frac{NM}{N_A\,a^3}
$$

Rearranging:

$$
a^3 = \frac{NM}{\rho N_A} \qquad\Rightarrow\qquad a = \left[\frac{NM}{\rho N_A}\right]^{1/3}
$$

Hence proved. For a simple cubic cell, $N = 8\times\frac{1}{8} = 1$; the same relation holds for BCC ($N = 2$) and FCC ($N = 4$) with the proper $N$.

#### Example
Copper (FCC): $N = 4$, $M = 63.55\times10^{-3}$ kg/mol, $\rho = 8960$ kg/m³:

$$
a = \left[\frac{4\times63.55\times10^{-3}}{8960\times6.022\times10^{23}}\right]^{1/3} = 3.61\times10^{-10}\ \text{m} = 0.361\ \text{nm}
$$

which agrees with the measured lattice constant of copper (0.3615 nm).

## Check yourself

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

1. Density of a crystal equals:
   A. Volume of cell / mass in cell
   B. Number of atoms / N_A
   C. Mass of atoms in one unit cell / volume of cell
   D. Atomic weight x N_A
   Answer: C. rho = m_cell/V_cell. (Key point: Density idea)

2. The mass of one atom is:
   A. M/N_A
   B. M x N_A
   C. N_A/M
   D. M/N
   Answer: A. One mole of mass M contains N_A atoms. (Key point: Mass of one atom)

3. For a cubic cell of side a, V_cell is:
   A. a^2
   B. 3a
   C. a^3
   D. a^3/N
   Answer: C. Volume of a cube is a^3. (Key point: Cell volume)

4. Rearranging rho = NM/(N_A a^3) gives a:
   A. (NM/rho N_A)^(1/3)
   B. NM/(rho N_A)
   C. (rho N_A/NM)^(1/3)
   D. (NM rho/N_A)^(1/3)
   Answer: A. a^3 = NM/(rho N_A), then take the cube root. (Key point: Result)

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