Chapter 1: Q95PP (page 1)

BIO “Seeing” Surfaces at the nanoscale. One technique for making images of surfaces at the nanometer scale, including membranes and biomolecules, is dynamic atomic force microscopy. In this technique, a small tip is attached to a cantilever, which is a flexible, rectangular slab supported at one end, like a diving board. The cantilever vibrates, so the tip moves up and down in simple harmonic motion. In one operating mode, the resonant frequency for a cantilever with force constant k = 1000 N/m is 100 kHz. As the oscillating tip is brought within a few nanometers of the surface of a sample (as shown in the figure), it experiences an attractive force from the surface. For an oscillation with a small amplitude (typically, 0.050 nm), the forceF that the sample surface exerts on the tip varies linearly with the displacement x of the tip, \(\left| F \right| = {k_{{\rm{surf}}}}x\), where \({k_{{\rm{surf}}}}\) is the effective force constant for this force. The net force on the tip is therefore\(\left( {k + {k_{{\rm{surf}}}}} \right)x\), and the frequency of the oscillation changes slightly due to the interaction with the surface. Measurements of the frequency as the tip moves over different parts of the sample’s surface can provide information about the sample.
In the model of Problem 14.94, what is the mechanical energy of the vibration when the tip is not interacting with the surface? (a) \({\bf{1}}.{\bf{2}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{18}}}}{\rm{ }}{\bf{J}}\); (b) \({\bf{1}}.{\bf{2}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{1}}6}}{\rm{ }}{\bf{J}}\);(c) \({\bf{1}}.{\bf{2}} \times {\bf{1}}{{\bf{0}}^{ - 9}}{\rm{ }}{\bf{J}}\); (d) \(5.0 \times {\bf{1}}{{\bf{0}}^{ - {\bf{8}}}}{\rm{ }}{\bf{J}}\).

Short Answer

Expert verified

Option (a) is correct.

Step by step solution

Given Data

\(\begin{array}{l}A = 0.050\;{\rm{nm}}\\k = 1000\;{\rm{N/m}}\\{\rm{f = }}\;{\rm{100}}\;{\rm{kHz}}\end{array}\)

Concept

The addition of kinetic energy and potential energy of an object is defined as the mechanical energy.

Determine the mechanical energy

Potential energy,

\(U = \frac{1}{2}k{x^2}\)

The potential energy is maximum at the maximum displacement from mean position and this potential energy is equal to the mechanical energy

Mechanical Energy,

\(\begin{array}{l}E = \frac{1}{2}k{A^2}\\A = {\rm{amplitude}}\end{array}\)

Substitute,

\(\begin{array}{l}A = 0.050\;{\rm{nm}}\\k = 1000\;{\rm{N/m}}\end{array}\)\(\begin{array}{c}E = \frac{1}{2}\left( {1000\;{\rm{N/m}}} \right){\left[ {\left( {0.050\;{\rm{nm}}} \right)\left( {\frac{{{{10}^{ - 9}}\;{\rm{m}}}}{{1\;{\rm{nm}}}}} \right)} \right]^2}\\ = 1.25 \times {10^{ - 18}}\;{\rm{J}}\end{array}\)

Hence Option (a) is correct.