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 m2 in cross section, would require an applied force F of 7.8 ×107 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 m2 cross section of gold is approximately 1/d2, 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:
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.
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 × 109 N m−2 (Pa).
From the question, l = 130µm, b = 7 µm and a = 10µm.
Using the stiffness formula (Equation 1.7):