# Question #96821

##### 1 Answer

#### Answer:

Here's how you can do that.

#### Explanation:

**Packing efficiency** is all about how much space is being occupied by the atoms present in a **unit cell**.

In order to calculate packing efficiency, you basically need to know three things

how manyatomsyou get per unit cellthe volume of asingle atomthetotal volumeof the unit cell

Packing efficiency will be equal to

#color(blue)(|bar(ul(color(white)(a/a)"pack. eff." = "volume occupied by atoms"/"total volume of the unit cell"color(white)(a/a)|)))#

So, start by calculating how many atoms you get in a **hexagonal closed-packed (HCP)** unit cell.

A HCP unit cell is a *hexegonal prism* that has a total of **lattice points**

three lattice pointsin the center of the celltwo lattice pointsin the centers of the basestwelve lattice pointsin the corners of the unit cell

Now, take a look at how the atoms are packed in the unit cell. Notice that you have

#1# atomfor every lattice point located in the center of the unit cell#1/2# of an atomfor every lattice point located in the center of the two bases#1/6"th"# of an atomfor every lattice point located in the corners of the unit cell

The total number of atoms that can fit in a HCP unit cell will thus be

#3 xx "1 atom" + 2 xx 1/2color(white)(a)"atoms" + 12 xx 1/6color(white)(a)"atoms" = "6 atoms"#

At this point, it would be easier to work with a *primitive unit cell*, which is equivalent to

Projecting this primitive cell here will get you

This primitive unit cell will contain **atoms**. If you take

#a = 2r implies r = a/2" " " "color(orange)("(*)")#

Here

Now, to get the volume of this primitive cell, you must use a *known* relationship that exists between the length of the cell and it *height*, usually given as ** not** derive this relationship here!

More specifically, you can use the fact that

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = sqrt(8/3) * a = (2sqrt(6))/3 * a)color(white)(a/a)|)))#

The volume of this primitive unit cell will thus be equal to the **area** of the rhombus that makes up its base and the height of the cell. Now the area of a rhombus can be expressed using the length of its side and **an interior angle**.

http://www.mathwords.com/a/area_rhombus.htm

#"area rhombus" = a^2 * sin(A)#

In your case, you can say that

#"area rhombus" = a^2 * sin(120^@) = sqrt(3)/2 * a^2#

**SIDE NOTE** *This will of course be equivalent to*

#"area rhombus" = a^2 * sin(60^@) = sqrt(3)/2 * a^2#

The volume of the primitive unit cell will thus be

#V_"primitive cell" = overbrace(sqrt(3)/color(red)(cancel(color(black)(2))) * a^2)^(color(green)("area of the rhombus")) xx overbrace((color(red)(cancel(color(black)(2)))sqrt(6))/3 * a)^(color(blue)("height of the cell"))#

#V_"primitive cell" = (sqrt(3) * sqrt(6))/3 * a^3 = sqrt(2) * a^3#

Since a primitive cell is equivalent to **total volume** of the unit cell will be

#V_"unit cell" = 3 xx sqrt(2) * a^3 = 3sqrt(2) * a^3#

The volume of a single atom is given by

#color(purple)(|bar(ul(color(white)(a/a)color(black)(V_"atom" = 4/3 * pi * r^3)color(white)(a/a)|)))#

Use equation

#V_"atom" = 4/3 * pi * (a/2)^3#

#V_"atom" = 4/3 * pi * a^3/8 = pi/6 * a^3#

Since you know that you get **atoms** in a unit cell, you can say that the **total occupied volume** will bee equal to

#V_"occupied" = color(red)(cancel(color(black)(6))) xx pi/color(red)(cancel(color(black)(6))) * a^3 = pi * a^3#

This means that the packing efficiency will be

#"pack. eff" = V_"occupied"/V_"unit cell"#

#"pack. eff" = (pi * color(red)(cancel(color(black)(a^3))))/(3sqrt(2) * color(red)(cancel(color(black)(a^3)))) = color(green)(|bar(ul(color(white)(a/a)0.7405color(white)(a/a)|)))#

You'll also see this expressed as *percent packing efficiency*

#"% pack. eff" = color(green)(|bar(ul(color(white)(a/a)74.05%color(white)(a/a)|)))#