# 3.6.1 Stiffness

Just how compliant does an AFM cantilever have to be to enable it to follow the undulations in a surface on an atomic scale? How can we find out? It turns out that this is easier than at first it seems.

A simple assumption we can make is that the compliance of the cantilever should be appreciably greater than that of a typical bond that holds atoms to one another. Here's one way in which a rough estimate of the stiffness (the force required to cause a given deflection) of the bonds in a solid can be made.

Take a solid material: gold, for example. Its Young's modulus, *E*, is 78 GPa. Using the equation relating stress σ, and strain ε (Hooke's law):

we can calculate that a piece of this material, 1 m^{2} in cross section, would require an applied force *F* of 7.8 ×10^{7} N to cause it to exhibit a strain of 0.1%.

X-ray diffraction measurements show the length, *d*, of the interatomic bonds in gold to be approximately 4 × 10^{−10} m, so a 0.1% strain corresponds to each bond that lies along the direction of the applied force changing in length by 4 × 10^{−13} m.

The number of bonds lying along the direction of the applied force in a 1 m^{2} cross section of gold is approximately 1/*d*^{2}, that is:

So, the force exerted on each bond is:

We have a force, and we have an extension caused by it. Dividing one by the other gives us the stiffness, *k*, of a gold–gold bond:

If the AFM probe is not to distort the surface as it travels across it, its stiffness should not be more than, say, 10% of that of the interatomic bond, giving us a maximum stiffness of about 3 N m^{−1}. For non-contact mode, somewhat stiffer cantilevers would be acceptable.

Now that we have an approximate value for the stiffness, how does that translate into the required dimensions for a cantilever? Let us assume, for simplicity, that the cantilever is a simple beam with a rectangular cross section (i.e. length *l*, width *a*, thickness *b*), rigidly anchored at one end. In this case, the standard expression for its deflection *z* at the far end under a point load *F*, as shown in Figure 12, is:

where *I* is the second moment of area of the cross section of the cantilever, given by:

## Example 1

If we chose to make our cantilever from a piece of aluminium kitchen foil of length *l* = 2 mm and thickness *b* = 10 µm (Young's modulus *E* = 70 GPa), what width would it need to be to have a stiffness of 3 N m^{−1}?

Stiffness, *k*, is force/deflection:

Therefore in terms of equations (1.4) and (1.5):

But we need to rearrange in terms of our unknown width *a*:

From this we find that *a* ~ 1.4 mm. These dimensions look easily achievable, almost by hand with a scalpel blade. To get a lower stiffness would just mean increasing the length. In practice, AFM cantilevers are manufactured with a range of stiffnesses from 0.01 N m^{−1} to 100 N m^{−1}, according to the modes in which they are intended to operate (the stiffer ones are used in the non-contact modes).

The aluminium foil cantilever is not a practical way of supporting an AFM tip. The tip itself needs to be fashioned using sub-micrometre precision. Furthermore, the tip needs to be attached to the cantilever. As we shall see in Section 3.7, this calls for a micro engineering solution using silicon or silicon nitride.

## SAQ 5

Estimate the stiffness of the 10 µm wide, 7 µm thick and 130 µm long cantilever beam that supports the AFM tip shown in Figure 10. It is made from material with E = 150 × 10^{9 }N m^{−2} (Pa).

### Answer

From the question, *l* = 130µm, *b* = 7 µm and *a* = 10µm.

Using the stiffness formula (Equation 1.7):