---
title: "Planar and linear density of BCC iron"
subject: "Material Science"
chapter: 2
key: "semester_3/material_science:planar-and-linear-density-bcc-iron"
source: https://www.bctengineer.com/q/semester_3/material_science/planar-and-linear-density-bcc-iron
site: BCT Engineering
---

# Planar and linear density of BCC iron

Material Science · Chapter 2: Atomic Structure, arrangement of atoms

- Asked in: year not given, 8 marks
- Video: [Linear and Planar Densities](https://www.youtube.com/watch?v=9Rd8349JwWc) — LearnChemE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/material_science/planar-and-linear-density-bcc-iron
- Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-2

## Question

Alpha iron is BCC with lattice constant a = 0.2866 nm. Calculate (a) the planar density of the (100) and (110) planes, (b) the linear density along the [100], [110] and [111] directions. Which plane and direction are close packed, and why does this matter for slip?

## Short answer

PD(100) = 12.17 atoms/nm², PD(110) = 17.22 atoms/nm². LD[100] = 3.49, [110] = 2.47, [111] = 4.03 atoms/nm. The close-packed plane is (110) and direction [111]; slip occurs easiest on these because they have highest atomic density and smallest Burgers vector.

## Answer

**Planar density** = atoms centred on the plane per unit area of the plane; **linear density** = atoms centred on the line per unit length.

#### (a) Planar density
**(100) plane:** a square of side $a$. Four corner atoms each contribute $\tfrac14$, giving 1 atom.

$$
\text{PD}_{(100)} = \frac{1}{a^2} = \frac{1}{(0.2866)^2} = 12.17\ \text{atoms/nm}^2
$$

**(110) plane:** a rectangle $a \times \sqrt{2}\,a$. Four corner atoms ($4\times\tfrac14 = 1$) plus the body-centre atom, which lies on this plane, give 2 atoms.

$$
\text{PD}_{(110)} = \frac{2}{\sqrt{2}\,a^2} = \frac{2}{1.4142 \times 0.08214} = 17.22\ \text{atoms/nm}^2
$$

#### (b) Linear density

| Direction | Length repeat | Atoms | Linear density (atoms/nm) |
|---|---|---|---|
| [100] | $a = 0.2866$ | 1 | $1/0.2866 = 3.49$ |
| [110] | $\sqrt2\,a = 0.4053$ | 1 | $1/0.4053 = 2.47$ |
| [111] | $\sqrt3\,a = 0.4964$ | 2 | $2/0.4964 = 4.03$ |

For [111], the body diagonal passes through two corner atoms and the centre atom, so 2 atoms are counted in the repeat length $\sqrt{3}\,a$.

#### Close-packed plane and direction
The densest plane is **(110)** (17.22 atoms/nm$^2$) and the densest direction is **[111]** (4.03 atoms/nm). Slip needs the least shear stress on close-packed planes and along close-packed directions, because the atomic spacing along such a direction is smallest (small Burgers vector) and such planes are the most widely spaced. Hence in BCC metals slip occurs on {110}<111> systems (and also on {112} and {123} planes).

**Answer: PD(100) = 12.17 atoms/nm², PD(110) = 17.22 atoms/nm²; LD[100] = 3.49, LD[110] = 2.47, LD[111] = 4.03 atoms/nm; close-packed plane (110), direction [111].**

## Check yourself

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

1. How many atoms lie on the BCC (110) plane rectangle a x sqrt(2)a?
   A. 1
   B. 2
   C. 4
   D. 3
   Answer: B. Four corners (4 x 1/4) plus the body-centre atom = 2. (Key point: Atoms on (110))

2. Which BCC plane has the greatest planar density?
   A. (100)
   B. (111)
   C. (110)
   D. (210)
   Answer: C. (110) has 17.22 atoms/nm^2, higher than (100) at 12.17. (Key point: Close-packed plane)

3. Which BCC direction is close packed and why?
   A. [100], with the shortest repeat
   B. [111], body diagonal through 2 atoms in repeat sqrt(3)a
   C. [110], face diagonal
   D. All equal
   Answer: B. [111] has linear density 4.03 atoms/nm, the highest. (Key point: Close-packed direction)

4. Why does slip occur on close-packed planes and directions?
   A. They need the least shear stress (small Burgers vector, widely spaced planes)
   B. They have the fewest atoms
   C. They have highest bond energy
   D. They are the only ones that exist
   Answer: A. Small atomic spacing along the direction and widely spaced planes make slip easiest; BCC slips on {110}<111>. (Key point: Slip systems)

## Related in Material Science

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- [BCC, FCC and HCP crystal structures](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Abcc-fcc-hcp-structures) — Material Science, chapter 2

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