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IPhO 2004, Теория — Задача 3. Atomic Probe Microscope

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Theoretical Question 3: Atomic Probe Microscope · 10 т.

Условие

Atomic probe microscopes (APMs) are powerful tools in the field of nano-science. The motion of a cantilever in APM can be detected by a photo-detector monitoring the reflected laser beam, as shown in Fig. 3.1. The cantilever can move only in the vertical direction and its displacement $z$ as a function of time $t$ can be described by the equation

m\frac{d^2z}{dt^2}+b\frac{dz}{dt}+kz=F, \qquad \text{(3.1)}

where $m$ is the cantilever mass, $k=m\omega_0^2$ is the spring constant of the cantilever, $b$ is a small damping coefficient satisfying $\omega_0 \gg (b/m)>0$ , and finally $F$ is an external driving force of the piezoelectric tube.

Schematic of a scanning probe microscope: a laser beam reflects off a cantilever (tip mass m, spring k) above a sample of stacked atoms, at height z=0, into a photo-detector whose output Vi=c2z goes to the input of a lock-in amplifier; a reference signal source VR feeds the lock-in amplifier and a phase shifter, whose output V'R drives the piezotube (force F) attached to the cantilever; the amplifier has an output line. An inset at lower right shows the mechanical model: piezotube, force F, spring k and tip mass m. — Figure 3.1 A schematic diagram for a scanning probe microscope (SPM). The inset in the lower right corner represents a simplified mechanical model to describe the coupling of the piezotube with the cantilever.

[Part A]

(a) [1.5 points] When $F=F_0\sin\omega t$ , $z(t)$ satisfying Eq. (3.1) can be written as $z(t)=A\sin(\omega t-\phi)$ , where $A>0$ and $0\le\phi\le\pi$ . Find the expression of the amplitude $A$ and $\tan\phi$ in terms of $F_0$ , $m$ , $\omega$ , $\omega_0$ , and $b$ . Obtain $A$ and the phase $\phi$ at the resonance frequency $\omega=\omega_0$ .

(b) [1 point] A lock-in amplifier shown in Fig.3.1 multiplies an input signal by the lock-in reference signal, $V_R=V_{R0}\sin\omega t$ , and then passes only the dc (direct current) component of the multiplied signal. Assume that the input signal is given by $V_i=V_{i0}\ \sin(\omega_i t-\phi_i)$ . Here $V_{R0}$ , $V_{i0}$ , $\omega_i$ , and $\phi_i$ are all positive given constants. Find the condition on $\omega$ ( $>0$ ) for a non-vanishing output signal. What is the expression for the magnitude of the non-vanishing dc output signal at this frequency?

(c) [1.5 points] Passing through the phase shifter, the lock-in reference voltage $V_R=V_{R0}\sin\omega t$ changes to $V'_R=V_{R0}\sin(\omega t+\pi/2)$ . $V'_R$ , applied to the piezoelectric tube, drives the cantilever with a force $F=c_1V'_R$ . Then, the photo-detector converts the displacement of the cantilever, $z$ , into a voltage $V_i=c_2z$ . Here $c_1$ and $c_2$ are constants. Find the expression for the magnitude of the dc output signal at $\omega=\omega_0$ .

(d) [2 points] The small change $\Delta m$ of the cantilever mass shifts the resonance frequency by $\Delta\omega_0$ . As a result, the phase $\phi$ at the original resonance frequency $\omega_0$ shifts by $\Delta\phi$ . Find the mass change $\Delta m$ corresponding to the phase shift $\Delta\phi=\pi/1800$ , which is a typical resolution in phase measurements. The physical parameters of the cantilever are given by $m=1.0\times10^{-12}\ \mathrm{kg}$ , $k=1.0\ \mathrm{N/m}$ , and $(b/m)=1.0\times10^{3}\ \mathrm{s^{-1}}$ . Use the approximations $(1+x)^a\approx1+ax$ and $\tan(\pi/2+x)\approx-1/x$ when $|x|\ll1$ .

[Part B]

From now on let us consider the situation that some forces, besides the driving force discussed in Part A, act on the cantilever due to the sample as shown in Fig.3.1.

(e) [1.5 points] Assuming that the additional force $f(h)$ depends only on the distance $h$ between the cantilever and the sample surface, one can find a new equilibrium position $h_0$ . Near $h=h_0$ , we can write $f(h)\approx f(h_0)+c_3(h-h_0)$ , where $c_3$ is a constant in $h$ . Find the new resonance frequency $\omega'_0$ in terms of $\omega_0$ , $m$ , and $c_3$ .

(f) [2.5 points] While scanning the surface by moving the sample horizontally, the tip of the cantilever charged with $Q=6e$ encounters an electron of charge $q=e$ trapped (localized in space) at some distance below the surface. During the scanning around the electron, the maximum shift of the resonance frequency $\Delta\omega_0\,(=\omega'_0-\omega_0)$ is observed to be much smaller than $\omega_0$ . Express the distance $d_0$ from the cantilever to the trapped electron at the maximum shift in terms of $m$ , $q$ , $Q$ , $\omega_0$ , $\Delta\omega_0$ , and the Coulomb constant $k_e$ . Evaluate $d_0$ in nm ( $1\ \mathrm{nm}=1\times10^{-9}\ \mathrm{m}$ ) for $\Delta\omega_0=20\ \mathrm{s^{-1}}$ .

The physical parameters of the cantilever are $m=1.0\times10^{-12}\ \mathrm{kg}$ and $k=1.0\ \mathrm{N/m}$ . Disregard any polarization effect in both the cantilever tip and the surface. Note that $k_e=1/4\pi\varepsilon_0=9.0\times10^{9}\ \mathrm{N\cdot m^2/C^2}$ and $e=-1.6\times10^{-19}\ \mathrm{C}$ .

Отговори

Решение

Theoretical Question 3: Scanning Probe Microscope

1. Answers

(a) $A=\dfrac{F_0}{\sqrt{m^2(\omega_0^2-\omega^2)^2+b^2\omega^2}}$ and $\tan\phi=\dfrac{b\omega_0}{m(\omega_0^2-\omega^2)}$ . At $\omega=\omega_0$ , $A=\dfrac{F_0}{b\omega_0}$ and $\phi=\dfrac{\pi}{2}$ .

(b) A non-vanishing dc component exists only when $\omega=\omega_i$ .

In this case the amplitude of the dc signal will be $\dfrac{1}{2}V_{i0}V_{R0}\cos\phi_i$ .

(d) $\Delta m=1.7\times10^{-18}\ \mathrm{kg}$ .

(e) $\omega'_0=\omega_0\left(1-\dfrac{c_3}{m\omega_0^2}\right)^{1/2}$ .

(f) $d_0=\left(k_e\dfrac{qQ}{m\omega_0\Delta\omega_0}\right)^{1/3}$

$d_0=41\ \mathrm{nm}$ .

2. Solutions

(a) [1.5 points]

Substituting $z(t)=A\sin(\omega t-\phi)$ in the equation $m\dfrac{d^2z}{dt^2}+b\dfrac{dz}{dt}+m\omega_0^2z=F_0\sin\omega t$ yields,

-m\omega^2\sin(\omega t-\phi)+b\omega\cos(\omega t-\phi)+m{\omega_0}^2\sin(\omega t-\phi)=\frac{F_0}{A}\sin\omega t. \qquad \text{(a1)}

Collecting terms proportional to $\sin\omega t$ and $\cos\omega t$ , one obtains

\left\{m(\omega_0^2-\omega^2)\cos\phi+b\omega\sin\phi-\frac{F_0}{A}\right\}\sin\omega t+\left\{-m(\omega_0^2-\omega^2)\sin\phi+b\omega\cos\phi\right\}\cos\omega t=0 \qquad \text{(a2)}

Zeroing the each curly square bracket produces

\tan\phi=\frac{b\omega}{m(\omega_0^2-\omega^2)}, \qquad \text{(a3)}

A=\frac{F_0}{\sqrt{m^2(\omega_0^2-\omega^2)^2+b^2\omega^2}}. \qquad \text{(a4)}

At $\omega=\omega_0$ ,

A=\frac{F_0}{b\omega_0}\ \text{ and }\ \phi=\frac{\pi}{2}. \qquad \text{(a5)}

(b) [1 point]

The multiplied signal is

\begin{aligned}&V_{i0}\sin(\omega_i t-\phi_i)V_{R0}\sin(\omega t)\\&=\frac{1}{2}V_{i0}V_{R0}[\cos\{(\omega_i-\omega)t-\phi_i\}-\cos\{(\omega_i+\omega)t-\phi_i\}]\end{aligned} \qquad \text{(b1)}

A non-vanishing dc component exists only when $\omega=\omega_i$ . In this case the amplitude of the dc signal will be

\frac{1}{2}V_{i0}V_{R0}\cos\phi_i. \qquad \text{(b2)}

(c) [1.5 points]

Since the lock-in amplifier measures the ac signal of the same frequency with its reference signal, the frequency of the piezoelectric tube oscillation, the frequency of the cantilever, and the frequency of the photodiode detector should be same. The magnitude of the input signal at the resonance is

V_{i0}=c_2\frac{F_0}{b\omega_0}=\frac{c_1c_2V_{R0}}{b\omega_0}. \qquad \text{(c1)}

Then, since the phase of the input signal is $-\dfrac{\pi}{2}+\dfrac{\pi}{2}=0$ at the resonance, $\phi_i=0$ and the lock-in amplifier signal is

\frac{1}{2}V_{i0}V_{R0}\cos0=\frac{c_1c_2}{2}\frac{V_{R0}^2}{b\omega_0}. \qquad \text{(c2)}

(d) [2 points]

The original resonance frequency $\omega_0=\sqrt{\dfrac{k}{m}}$ is shifted to

\sqrt{\frac{k}{m+\Delta m}}=\sqrt{\frac{k}{m}}\left(1+\frac{\Delta m}{m}\right)^{-\frac{1}{2}}\cong\sqrt{\frac{k}{m}}\left(1-\frac{1}{2}\frac{\Delta m}{m}\right)=\omega_0\left(1-\frac{1}{2}\frac{\Delta m}{m}\right). \qquad \text{(d1)}

Thus

\Delta\omega_0=-\frac{1}{2}\omega_0\frac{\Delta m}{m}. \qquad \text{(d2)}

Near the resonance, by substituting $\phi\to\dfrac{\pi}{2}+\Delta\phi$ and $\omega_0\to\omega_0+\Delta\omega_0$ in Eq. (a3), the change of the phase due to the small change of $\omega_0$ (not the change of $\omega$ ) is

\tan\left(\frac{\pi}{2}+\Delta\phi\right)=-\frac{1}{\tan\Delta\phi}=\frac{b}{2m\Delta\omega_0}. \qquad \text{(d3)}

Therefore,

\Delta\phi\approx\tan\Delta\phi=-\frac{2m\Delta\omega_0}{b}. \qquad \text{(d4)}

From Eqs. (d2) and (d4),

\Delta m=\frac{b}{\omega_0}\Delta\phi=\frac{10^3\cdot10^{-12}}{10^6}\frac{\pi}{1800}=\frac{\pi}{1.8}10^{-18}=1.7\times10^{-18}\ \mathrm{kg}. \qquad \text{(d5)}

(e) [1.5 points]

In the presence of interaction, the equation of motion near the new equilibrium position $h_0$ becomes

m\frac{d^2z}{dt^2}+b\frac{dz}{dt}+m\omega_0^2z-c_3z=F_0\sin\omega t \qquad \text{(e1)}

where we used $f(h)\approx f(h_0)+c_3z$ with $z=h-h_0$ being the displacement from the new equilibrium position $h_0$ . Note that the constant term $f(h_0)$ is cancelled at the new equilibrium position.

Thus the original resonance frequency $\omega_0=\sqrt{\dfrac{k}{m}}$ will be shifted to

\omega'_0=\sqrt{\frac{k-c_3}{m}}=\sqrt{\frac{m\omega_0^2-c_3}{m}}=\omega_0\sqrt{1-\frac{c_3}{m\omega_0^2}}. \qquad \text{(e3)}

Hence the resonance frequency shift is given by

\Delta\omega_0=\omega_0\left[\sqrt{1-\frac{c_3}{m\omega_0^2}}-1\right]. \qquad \text{(e4)}

(f) [2.5 points]

The maximum shift occurs when the cantilever is on top of the charge, where the interacting force is given by

f(h)=k_e\frac{qQ}{h^2}. \qquad \text{(f1)}

From this,

c_3=\left.\frac{df}{dh}\right|_{h=d_0}=-2k_e\frac{qQ}{d_0^3}. \qquad \text{(f2)}

Since $\Delta\omega_0\ll\omega_0$ , we can approximate Eq. (e4) as

\Delta\omega_0\approx-\frac{c_3}{2m\omega_0}. \qquad \text{(f3)}

From Eqs. (f2) and (f3), we have

\Delta\omega_0=-\frac{1}{2m\omega_0}\left(-2k_e\frac{qQ}{d_0^3}\right)=k_e\frac{qQ}{m\omega_0d_0^3}. \qquad \text{(f4)}

Here $q=e=-1.6\times10^{-19}$ Coulomb and $Q=6e=-9.6\times10^{-19}$ Coulomb. Using the values provided,

d_0=\left(k_e\frac{qQ}{m\omega_0\Delta\omega_0}\right)^{1/3}=4.1\times10^{-8}\ \mathrm{m}=41\ \mathrm{nm}. \qquad \text{(f5)}

Thus the trapped electron is 41 nm from the cantilever.

3. Mark Distribution

No.	Total Pt.	Partial Pt.	Contents
(a)	1.5	0.7	Equations for $A$ and $\phi$ (substitution and manipulation)
		0.4	Correct answers for $A$ and $\phi$
		0.4	$A$ and $\phi$ at $\omega_0$
(b)	1.0	0.4	Equation for the multiplied signal
		0.3	Condition for the non-vanishing dc output
		0.3	Correct answer for the dc output
(c)	1.5	0.6	Relation between $V_i$ and $V_R$
		0.4	Condition for the maximum dc output
		0.5	Correct answer for the magnitude of dc output
(d)	2.0	0.5	Relation between $\Delta m$ and $\Delta\omega_0$
		1.0	Relations between $\Delta\omega_0$ (or $\Delta m$ ) and $\Delta\phi$
		0.5	Correct answer (Partial credit of 0.2 for the wrong sign.)
(e)	1.5	1.0	Modification of the equation with $f(h)$ and use of a proper approximation for the equation
		0.5	Correct answer
(f)	2.5	0.5	Use of a correct formula of Coulomb force
		0.3	Evaluation of $c_3$
		0.6	Use of the result in (e) for either $\Delta\omega_0$ or $\omega'^2_0-\omega^2_0$
		0.6	Expression for $d_0$
		0.5	Correct answer
Total	10

Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf