IPhO 2013 · Theory problem 2

Plasmonic Steam Generator

Introduction

In this problem we study an efficient process of steam production that has been demonstrated to work experimentally. An aqueous solution of spherical nanometer-sized silver spheres (nanoparticles) with only about \(10^{13}\) particles per liter is illuminated by a focused light beam. A fraction of the light is absorbed by the nanoparticles, which are heated up and generate steam locally around them without heating up the entire water solution. The steam is released from the system in the form of escaping steam bubbles. Not all details of the process are well understood at present, but the core process is known to be absorption of light through the so-called collective electron oscillations of the metallic nanoparticles. The device is known as a plasmonic steam generator.

Figure 1

Figure 2.1 (a) A spherical charge-neutral nanoparticle of radius \(R\) placed at the center of the coordinate system. (b) A sphere with a positive homogeneous charge density \(\rho\) (red), and containing a smaller spherical charge-neutral region (0, yellow) of radius \(R_1\), with its center displaced by \(\boldsymbol{x}_{\mathrm d}=x_{\mathrm d}\boldsymbol{e}_x\). (c) The sphere with positive charge density \(\rho\) of the nanoparticle silver ions is fixed in the center of the coordinate system. The center of the spherical region with negative spherical charge density \(-\rho\) (blue) of the electron cloud is displaced by \(\boldsymbol{x}_{\mathrm p}\), where \(x_{\mathrm p}\ll R\). (d) An external homogeneous electric field \(\boldsymbol{E}_0=-E_0\boldsymbol{e}_x\). For time-dependent \(\boldsymbol{E}_0\), the electron cloud moves with velocity \(\boldsymbol{v}=\mathrm d\boldsymbol{x}_{\mathrm p}/\mathrm dt\). (e) The rectangular vessel \((h\times h\times a)\) containing the aqueous solution of nanoparticles illuminated by monochromatic light propagating along the \(z\)-axis with angular frequency \(\omega_{\mathrm p}\) and intensity \(S\).

A single spherical silver nanoparticle

Throughout this problem we consider a spherical silver nanoparticle of radius \(R=10{,}0\ \mathrm{nm}\) and with its center fixed at the origin of the coordinate system, see Fig. 2.1(a). All motions, forces and driving fields are parallel to the horizontal \(x\)-axis (with unit vector \(\boldsymbol{e}_x\)). The nanoparticle contains free (conduction) electrons moving within the whole nanoparticle volume without being bound to any silver atom. Each silver atom is a positive ion that has donated one such free electron.

Part Question Points
2.1 Find the following quantities: The volume \(V\) and mass \(M\) of the nanoparticle, the number \(N\) and charge density \(\rho\) of silver ions in the particle, and for the free electrons their concentration \(n\), their total charge \(Q\), and their total mass \(m_0\). 0.7

The electric field in a charge-neutral region inside a charged sphere

For the rest of the problem assume that the relative dielectric permittivity of all materials is \(\varepsilon=1\). Inside a charged sphere of homogeneous charge density \(\rho\) and radius \(R\) is created a small spherical charge-neutral region of radius \(R_1\) by adding the opposite charge density \(-\rho\), with its center displaced by \(\boldsymbol{x}_{\mathrm d}=x_{\mathrm d}\boldsymbol{e}_x\) from the center of the \(R\)-sphere, see Fig. 2.1(b).

Part Question Points
2.2 Show that the electric field inside the charge-neutral region is homogenous of the form \(\boldsymbol{E}=A\,(\rho/\varepsilon_0)\,\boldsymbol{x}_{\mathrm d}\), and determine the pre-factor \(A\). 1.2

The restoring force on the displaced electron cloud

In the following, we study the collective motion of the free electrons, and therefore model them as a single negatively charged sphere of homogeneous charge density \(-\rho\) with a center position \(\boldsymbol{x}_{\mathrm p}\), which can move along the \(x\)-axis relative to the center of the positively charged sphere (silver ions) fixed at the origin of the coordinate system, see Fig. 2.1(c). Assume that an external force \(\boldsymbol{F}_{\mathrm{ext}}\) displaces the electron cloud to a new equilibrium position \(\boldsymbol{x}_{\mathrm p}=x_{\mathrm p}\boldsymbol{e}_x\) with \(\lvert x_{\mathrm p}\rvert\ll R\). Except for tiny net charges at opposite ends of the nanoparticle, most of its interior remains charge-neutral.

Part Question Points
2.3 Express in terms of \(\boldsymbol{x}_{\mathrm p}\) and \(n\) the following two quantities: The restoring force \(\boldsymbol{F}\) exerted on the electron cloud and the work \(W_{\mathrm{el}}\) done on the electron cloud during displacement. 1.0

The spherical silver nanoparticle in an external constant electric field

A nanoparticle is placed in vacuum and influenced by an external force \(\boldsymbol{F}_{\mathrm{ext}}\) due to an applied static homogeneous electric field \(\boldsymbol{E}_0=-E_0\boldsymbol{e}_x\), which displaces the electron cloud the small distance \(\lvert x_{\mathrm p}\rvert\), where \(\lvert x_{\mathrm p}\rvert\ll R\).

Part Question Points
2.4 Find the displacement \(x_{\mathrm p}\) of the electron cloud in terms of \(E_0\) and \(n\), and determine the amount \(-\Delta Q\) of electron charge displaced through the \(yz\)-plane at the center of the nanoparticle in terms of \(n\), \(R\) and \(x_{\mathrm p}\). 0.6

The equivalent capacitance and inductance of the silver nanoparticle

For both a constant and a time-dependent field \(\boldsymbol{E}_0\), the nanoparticle can be modeled as an equivalent electric circuit. The equivalent capacitance can be found by relating the work \(W_{\mathrm{el}}\) done on the separation of charges \(\Delta Q\), to the energy of a capacitor, carrying charge \(\pm\Delta Q\). The charge separation will cause a certain equivalent voltage \(V_0\) across the equivalent capacitor.

Part Question Points
2.5a Express the systems equivalent capacitance \(C\) in terms of \(\varepsilon_0\) and \(R\), and find its value. 0.7
2.5b For this capacitance, determine in terms of \(E_0\) and \(R\) the equivalent voltage \(V_0\) that should be connected to the equivalent capacitor in order to accumulate the charge \(\Delta Q\). 0.4

For a time-dependent field \(\boldsymbol{E}_0\), the electron cloud moves with velocity \(\boldsymbol{v}=v\boldsymbol{e}_x\), Fig. 2.1(d). It has the kinetic energy \(W_{\mathrm{kin}}\) and forms an electric current \(I\) flowing through the fixed \(yz\)-plane. The kinetic energy of the electron cloud can be attributed to the energy of an equivalent inductor of inductance \(L\) carrying the current \(I\).

Part Question Points
2.6a Express both \(W_{\mathrm{kin}}\) and \(I\) in terms of the velocity \(v\). 0.7
2.6b Express the equivalent inductance \(L\) in terms of the particle radius \(R\), the electron charge \(e\) and mass \(m_e\), the electron concentration \(n\), and calculate its value. 0.5

The plasmon resonance of the silver nanoparticle

From the above analysis it follows that the motion, arising from displacing the electron cloud from its equilibrium position and then releasing it, can be modeled by an ideal \(LC\)-circuit oscillating at resonance. This dynamical mode of the electron cloud is known as the plasmon resonance, which oscillates at the so-called angular plasmon frequency \(\omega_{\mathrm p}\).

Part Question Points
2.7a Find an expression for the angular plasmon frequency \(\omega_{\mathrm p}\) of the electron cloud in terms of the electron charge \(e\) and mass \(m_e\), the electron density \(n\), and the permittivity \(\varepsilon_0\). 0.5
2.7b Calculate \(\omega_{\mathrm p}\) in rad/s and the wavelength \(\lambda_{\mathrm p}\) in nm of light in vacuum having angular frequency \(\omega=\omega_{\mathrm p}\). 0.4

The silver nanoparticle illuminated with light at the plasmon frequency

In the rest of the problem, the nanoparticle is illuminated by monochromatic light at the angular plasmon frequency \(\omega_{\mathrm p}\) with the incident intensity \(S=\frac12c\varepsilon_0E_0^2=1.00\ \mathrm{MW\,m^{-2}}\). As the wavelength is large, \(\lambda_{\mathrm p}\gg R\), the nanoparticle can be considered as being placed in a homogeneous harmonically oscillating field \(\boldsymbol{E}_0=-E_0\cos(\omega_{\mathrm p}t)\boldsymbol{e}_x\). Driven by \(\boldsymbol{E}_0\), the center \(\boldsymbol{x}_{\mathrm p}(t)\) of the electron cloud oscillates at the same frequency with velocity \(\boldsymbol{v}=\mathrm d\boldsymbol{x}_{\mathrm p}/\mathrm dt\) and constant amplitude \(x_0\). This oscillating electron motion leads to absorption of light. The energy captured by the particle is either converted into Joule heating inside the particle or re-emitted by the particle as scattered light.

Joule heating is caused by random inelastic collisions, where any given free electron once in a while hits a silver ion and loses its total kinetic energy, which is converted into vibrations of the silver ions (heat). The average time between the collisions is \(\tau\gg1/\omega_{\mathrm p}\), where for silver nanoparticle we use \(\tau=5.24\times10^{-15}\ \mathrm s\).

Part Question Points
2.8a Find an expression for the time-averaged Joule heating power \(P_{\mathrm{heat}}\) in the nanoparticle as well as the time-averaged current squared \(\langle I^2\rangle\), which includes explicitly the time-averaged velocity squared \(\langle v^2\rangle\) of the electron cloud. 1.0
2.8b Find an expression for the equivalent ohmic resistance \(R_{\mathrm{heat}}\) in an equivalent resistor-model of the nanoparticle having the Joule heating power \(P_{\mathrm{heat}}\) due to the electron cloud current \(I\). Calculate the numerical value of \(R_{\mathrm{heat}}\). 1.0

The incident light beam loses some time-averaged power \(P_{\mathrm{scat}}\) by scattering on the oscillating electron cloud (re-emission). \(P_{\mathrm{scat}}\) depends on the scattering source amplitude \(x_0\), charge \(Q\), angular frequency \(\omega_{\mathrm p}\) and properties of the light (the speed of light \(c\) and permittivity \(\varepsilon_0\) in vacuum). In terms of these four variables, \(P_{\mathrm{scat}}\) is given by \[ P_{\mathrm{scat}}=\frac{Q^2x_0^2\omega_{\mathrm p}^4}{12\pi\varepsilon_0c^3}. \]

Part Question Points
2.9 By use of \(P_{\mathrm{scat}}\), find an expression for the equivalent scattering resistance \(R_{\mathrm{scat}}\) (in analogy with \(R_{\mathrm{heat}}\)) in an equivalent resistor-model, and calculate its value. 1.0

The above equivalent circuit elements are combined into an \(LCR\) series circuit model of the silver nanoparticle, which is driven by a harmonically oscillating equivalent voltage \(V=V_0\cos(\omega_{\mathrm p}t)\) determined by the electric field \(\boldsymbol{E}_0\) of the incident light.

Part Question Points
2.10a Derive expressions for the time-averaged power losses \(P_{\mathrm{heat}}\) and \(P_{\mathrm{scat}}\) involving the amplitude \(E_0\) of the electric field in the incident light at the plasmon resonance \(\omega=\omega_{\mathrm p}\). 1.2
2.10b Calculate the numerical value of \(E_0\), \(P_{\mathrm{heat}}\), and \(P_{\mathrm{scat}}\). 0.3

Steam generation by light

An aqueous solution of silver nanoparticles is prepared with a concentration \(n_{\mathrm{np}}=7.3\times10^{15}\ \mathrm{m^{-3}}\). It is placed inside a rectangular transparent vessel of size \(h\times h\times a=10\times10\times1.0\ \mathrm{cm^3}\) and illuminated by light at the plasmon frequency with the same intensity \(S=1.00\ \mathrm{MW\,m^{-2}}\) at normal incidence as above, see Fig. 2.1(e). The temperature of the water is \(T_{\mathrm{wa}}=20\,^\circ\mathrm C\) and we assume, in fair agreement with observations, that in steady state all Joule heating of the nanoparticle goes to the production of steam of temperature \(T_{\mathrm{st}}=110\,^\circ\mathrm C\), without raising the temperature of the water.

The thermodynamic efficiency \(\eta\) of the plasmonic steam generator is defined by the power ratio \(\eta=P_{\mathrm{st}}/P_{\mathrm{tot}}\), where \(P_{\mathrm{st}}\) is the power going into the production of steam in the entire vessel, while \(P_{\mathrm{tot}}\) is the total power of the incoming light that enters the vessel.

Most of the time any given nanoparticle is surrounded by steam instead of water, and it can thus be described as being in vacuum.

Part Question Points
2.11a Calculate the total mass per second \(\mu_{\mathrm{st}}\) of steam produced by the plasmonic steam generator during illumination by light at the plasmon frequency and intensity \(S\). 0.6
2.11b Calculate the numerical value of the thermodynamic efficiency \(\eta\) of the plasmonic steam generator. 0.2

Data sheet: Table of physical parameters

Physical parameter Value
Speed of light in vacuum \(c=2.998\times10^8\ \mathrm{m\,s^{-1}}\)
Planck's constant over \(2\pi\) \(\hbar=1.055\times10^{-34}\ \mathrm{J\,s}\)
Gravitational constant \(G=6.67\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}\)
Gravitational acceleration \(g=9.82\ \mathrm{m\,s^{-2}}\)
Elementary charge \(e=1.602\times10^{-19}\ \mathrm C\)
Electric permittivity of vacuum \(\varepsilon_0=8.854\times10^{-12}\ \mathrm{C^2\,J^{-1}\,m^{-1}}\)
Electron mass \(m_e=9.109\times10^{-31}\ \mathrm{kg}\)
Avogadro constant \(N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}\)
Boltzmann constant \(k_{\mathrm B}=1.381\times10^{-23}\ \mathrm{J\,K^{-1}}\)
Stony meteorite, specific heat \(c_{\mathrm{sm}}=1.2\times10^3\ \mathrm{J\,kg^{-1}\,K^{-1}}\)
Stony meteorite, thermal conductivity \(k_{\mathrm{sm}}=2.0\ \mathrm{W\,m^{-1}\,K^{-1}}\)
Stony meteorite, density \(\rho_{\mathrm{sm}}=3.3\times10^3\ \mathrm{kg\,m^{-3}}\)
Stony meteorite, melting point \(T_{\mathrm{sm}}=1.7\times10^3\ \mathrm K\)
Stony meteorite, specific melting heat \(L_{\mathrm{sm}}=2.6\times10^5\ \mathrm{J\,kg^{-1}}\)
Silver, molar mass \(M_{\mathrm{Ag}}=1.079\times10^{-1}\ \mathrm{kg\,mol^{-1}}\)
Silver, density \(\rho_{\mathrm{Ag}}=1.049\times10^4\ \mathrm{kg\,m^{-3}}\)
Silver, specific heat capacity \(c_{\mathrm{Ag}}=2.40\times10^2\ \mathrm{J\,kg^{-1}\,K^{-1}}\)
Water, molar mass \(M_{\mathrm{wa}}=1.801\times10^{-2}\ \mathrm{kg\,mol^{-1}}\)
Water, density \(\rho_{\mathrm{wa}}=0.998\times10^3\ \mathrm{kg\,m^{-3}}\)
Water, specific heat capacity \(c_{\mathrm{wa}}=4.181\times10^3\ \mathrm{J\,kg^{-1}\,K^{-1}}\)
Water, heat of vaporization \(L_{\mathrm{wa}}=2.260\times10^6\ \mathrm{J\,kg^{-1}}\)
Water, boiling temperature \(T_{100}=100\,^\circ\mathrm C=373.15\ \mathrm K\)
Ice, density of glacier \(\rho_{\mathrm{ice}}=0.917\times10^3\ \mathrm{kg\,m^{-3}}\)
Steam, specific heat capacity \(c_{\mathrm{st}}=2.080\times10^3\ \mathrm{J\,kg^{-1}\,K^{-1}}\)
Earth, mass of the \(m_{\mathrm E}=5.97\times10^{24}\ \mathrm{kg}\)
Earth, radius of the \(R_{\mathrm E}=6.38\times10^6\ \mathrm m\)
Sun, mass of the \(m_{\mathrm S}=1.99\times10^{30}\ \mathrm{kg}\)
Sun, radius of the \(R_{\mathrm S}=6.96\times10^8\ \mathrm m\)
Average Sun-Earth distance \(a_{\mathrm E}=1.50\times10^{11}\ \mathrm m\)

Official English text of the problem.

Answers

Here \(e>0\), \(Q<0\), and \(W_{\mathrm{el}}\) denotes the positive work supplied to separate the charges. Choose the circuit-current direction so that \(I=\mathrm d(\Delta Q)/\mathrm dt\), positive when the electrons move in \(+x\).

  • 2.1 \[ \begin{gathered} V=4.19\times10^{-24}\ \mathrm{m^3},\qquad M=4.39\times10^{-20}\ \mathrm{kg},\\ N=2.45\times10^5,\qquad \rho=+9.38\times10^9\ \mathrm{C\,m^{-3}},\\ n=5.85\times10^{28}\ \mathrm{m^{-3}},\qquad Q=-3.93\times10^{-14}\ \mathrm C,\qquad m_0=2.23\times10^{-25}\ \mathrm{kg}. \end{gathered} \]

  • 2.2 \[ \boldsymbol E=\frac{\rho}{3\varepsilon_0}\boldsymbol x_{\mathrm d}, \qquad \boxed{A=\frac13}. \]

  • 2.3 \[ \boldsymbol F=-\frac{4\pi n^2e^2R^3}{9\varepsilon_0}\boldsymbol x_{\mathrm p}, \qquad W_{\mathrm{el}}=\frac{2\pi n^2e^2R^3}{9\varepsilon_0}x_{\mathrm p}^2. \]

  • 2.4 \[ x_{\mathrm p}=\frac{3\varepsilon_0E_0}{ne}, \qquad -\Delta Q=-ne\pi R^2x_{\mathrm p}. \]

  • 2.5a \[ C=\frac94\pi\varepsilon_0R =6.26\times10^{-19}\ \mathrm F. \]

  • 2.5b \[ V_0=\frac43RE_0. \]

  • 2.6a \[ W_{\mathrm{kin}}=\frac12m_0v^2=\frac{2\pi}{3}nm_eR^3v^2, \qquad I=ne\pi R^2v. \]

  • 2.6b \[ L=\frac{4m_e}{3\pi ne^2R} =2.57\times10^{-14}\ \mathrm H. \]

  • 2.7a \[ \omega_{\mathrm p}=\sqrt{\frac{ne^2}{3\varepsilon_0m_e}}. \]

  • 2.7b \[ \omega_{\mathrm p}=7.88\times10^{15}\ \mathrm{rad\,s^{-1}}, \qquad \lambda_{\mathrm p}=239\ \mathrm{nm}. \]

  • 2.8a \[ P_{\mathrm{heat}}=\frac{m_0}{2\tau}\langle v^2\rangle, \qquad \langle I^2\rangle=(ne\pi R^2)^2\langle v^2\rangle. \]

  • 2.8b \[ R_{\mathrm{heat}}=\frac{L}{2\tau} =\frac{2m_e}{3\pi ne^2R\tau} =2.46\ \Omega. \]

  • 2.9 \[ R_{\mathrm{scat}}=\frac{8R^2\omega_{\mathrm p}^2} {27\pi\varepsilon_0c^3} =2.45\ \Omega. \]

  • 2.10a \[ P_{\mathrm{heat}}= \frac{8R^2E_0^2}{9} \frac{R_{\mathrm{heat}}}{(R_{\mathrm{heat}}+R_{\mathrm{scat}})^2}, \qquad P_{\mathrm{scat}}= \frac{8R^2E_0^2}{9} \frac{R_{\mathrm{scat}}}{(R_{\mathrm{heat}}+R_{\mathrm{scat}})^2}. \]

  • 2.10b \[ E_0=2.74\times10^4\ \mathrm{V\,m^{-1}}, \qquad P_{\mathrm{heat}}=6.82\times10^{-9}\ \mathrm W, \qquad P_{\mathrm{scat}}=6.82\times10^{-9}\ \mathrm W. \]

  • 2.11a, using uniform illumination throughout the vessel: \[ \mu_{\mathrm{st}}\simeq1.9\times10^{-3}\ \mathrm{kg\,s^{-1}} =1.9\ \mathrm{g\,s^{-1}}. \]

  • 2.11b, with the same approximation: \[ \eta\simeq0.50=50\%. \]

Solution

All electrostatic and circuit results below use the prescribed rigid, uniformly charged electron-cloud model. Displacement-dependent results retain the leading terms for \(\lvert x_{\mathrm p}\rvert/R\ll1\).

2.1

The particle is a sphere, so \[ V=\frac{4\pi R^3}{3} =\frac{4\pi}{3}(1.00\times10^{-8})^3 =4.19\times10^{-24}\ \mathrm{m^3}. \] Its mass is \[ M=\rho_{\mathrm{Ag}}V =4.39\times10^{-20}\ \mathrm{kg}. \]

The number of silver atoms, and hence of singly charged silver ions, is \[ N=\frac{M}{M_{\mathrm{Ag}}}N_A =2.45\times10^5. \] There is one conduction electron per ion. Thus \[ n=\frac{N}{V} =\frac{\rho_{\mathrm{Ag}}N_A}{M_{\mathrm{Ag}}} =5.85\times10^{28}\ \mathrm{m^{-3}}. \]

The positive ionic charge density is \[ \rho=ne=+9.38\times10^9\ \mathrm{C\,m^{-3}}. \] For the electrons, \[ Q=-Ne=-3.93\times10^{-14}\ \mathrm C, \qquad m_0=Nm_e=2.23\times10^{-25}\ \mathrm{kg}. \]

2.2

Apply Gauss's law to a concentric Gaussian sphere of radius \(r\) inside a uniformly charged sphere: \[ E(r)\,4\pi r^2 =\frac{\rho(4\pi r^3/3)}{\varepsilon_0}. \] Consequently, the interior field is \[ \boldsymbol E_+(\boldsymbol r) =\frac{\rho}{3\varepsilon_0}\boldsymbol r. \]

The neutral region is formed by superposing a sphere of charge density \(-\rho\), centered at \(\boldsymbol x_{\mathrm d}\). At a point inside this smaller sphere its field is \[ \boldsymbol E_-(\boldsymbol r) =-\frac{\rho}{3\varepsilon_0} (\boldsymbol r-\boldsymbol x_{\mathrm d}). \] Adding the two fields gives \[ \boldsymbol E =\boldsymbol E_++\boldsymbol E_- =\frac{\rho}{3\varepsilon_0}\boldsymbol x_{\mathrm d}. \] The position \(\boldsymbol r\) cancels, so the field is homogeneous and \[ \boxed{A=\frac13}. \]

2.3

The displaced electron cloud has charge density \(-ne\), while the fixed ions have charge density \(+ne\). By 2.2, the field in their overlapping, neutral interior is \[ \boldsymbol E_{\mathrm{int}} =\frac{ne}{3\varepsilon_0}\boldsymbol x_{\mathrm p}. \] To first order in displacement this field gives the restoring force on the cloud: \[ \boldsymbol F=Q\boldsymbol E_{\mathrm{int}} =-\frac{Nne^2}{3\varepsilon_0}\boldsymbol x_{\mathrm p}. \] The small nonoverlapping regions only affect higher-order terms. With \(N=4\pi nR^3/3\), \[ \boxed{\boldsymbol F =-\frac{4\pi n^2e^2R^3}{9\varepsilon_0}\boldsymbol x_{\mathrm p}}. \]

Define the effective spring constant \[ k=\frac{Nne^2}{3\varepsilon_0}. \] The work supplied in a quasistatic displacement from zero to \(x_{\mathrm p}\) is \[ W_{\mathrm{el}} =-\int_0^{x_{\mathrm p}}F_x(x)\,\mathrm dx =\int_0^{x_{\mathrm p}}kx\,\mathrm dx =\frac12kx_{\mathrm p}^2. \] Therefore \[ \boxed{W_{\mathrm{el}} =\frac{Nne^2}{6\varepsilon_0}x_{\mathrm p}^2 =\frac{2\pi n^2e^2R^3}{9\varepsilon_0}x_{\mathrm p}^2}. \] This is the stored electrostatic energy used in 2.5. The work done by the restoring force itself is \(-W_{\mathrm{el}}\).

2.4

The external field points in \(-x\), so the force on the negative electron cloud points in \(+x\): \[ \boldsymbol F_{\mathrm{ext}} =Q\boldsymbol E_0 =NeE_0\boldsymbol e_x. \] At equilibrium, \[ NeE_0-\frac{Nne^2}{3\varepsilon_0}x_{\mathrm p}=0. \] Hence \[ \boxed{x_{\mathrm p}=\frac{3\varepsilon_0E_0}{ne}}. \]

For a small translation \(x_{\mathrm p}\), the volume of electron cloud transported through the central plane is, to leading order, \[ \Delta V=\pi R^2x_{\mathrm p}. \] Multiplying by its charge density gives \[ \boxed{-\Delta Q=-ne\pi R^2x_{\mathrm p}}. \] Thus the positive charge-separation variable is \[ \Delta Q=ne\pi R^2x_{\mathrm p} =3\pi\varepsilon_0R^2E_0. \]

2.5a

Equate the electrostatic energy to the energy of the equivalent capacitor: \[ W_{\mathrm{el}}=\frac{(\Delta Q)^2}{2C}. \] Using the results above, \[ C= \frac{(ne\pi R^2x_{\mathrm p})^2} {2\left(2\pi n^2e^2R^3x_{\mathrm p}^2/(9\varepsilon_0)\right)} =\frac94\pi\varepsilon_0R. \] Therefore \[ \boxed{C=6.26\times10^{-19}\ \mathrm F}. \]

2.5b

The equivalent voltage is \[ V_0=\frac{\Delta Q}{C}. \] Substitution yields \[ V_0= \frac{3\pi\varepsilon_0R^2E_0} {(9/4)\pi\varepsilon_0R}, \] so \[ \boxed{V_0=\frac43RE_0}. \]

2.6a

All \(N\) electrons participate in the collective motion. Their kinetic energy is \[ \boxed{ W_{\mathrm{kin}}=\frac12Nm_ev^2 =\frac12m_0v^2 =\frac{2\pi}{3}nm_eR^3v^2}. \]

Differentiating the charge-separation relation gives \[ \boxed{ I=\frac{\mathrm d(\Delta Q)}{\mathrm dt} =ne\pi R^2v}. \] Here positive circuit current corresponds to conventional current in the \(-x\) direction. If conventional current through the plane were instead defined positive in \(+x\), it would be \(I_x=-ne\pi R^2v\). The energy and resistance results depend on \(I^2\), so this orientation choice does not affect them.

2.6b

Identify the kinetic energy with the energy of the equivalent inductor: \[ \frac12LI^2=\frac12m_0v^2. \] It follows that \[ L=\frac{m_0}{(ne\pi R^2)^2}. \] Using \(m_0=(4\pi R^3/3)nm_e\), \[ \boxed{L=\frac{4m_e}{3\pi ne^2R} =2.57\times10^{-14}\ \mathrm H}. \]

2.7a

For a freely oscillating ideal \(LC\) circuit, \[ L\frac{\mathrm d^2(\Delta Q)}{\mathrm dt^2} +\frac{\Delta Q}{C}=0, \] so \[ \omega_{\mathrm p}^2=\frac1{LC}. \] Since \[ LC= \frac{4m_e}{3\pi ne^2R} \frac{9\pi\varepsilon_0R}{4} =\frac{3\varepsilon_0m_e}{ne^2}, \] we obtain \[ \boxed{\omega_{\mathrm p} =\sqrt{\frac{ne^2}{3\varepsilon_0m_e}}}. \] The same result follows directly from \(\omega_{\mathrm p}^2=k/m_0\).

2.7b

Substitution of the given constants gives \[ \boxed{\omega_{\mathrm p} =7.88\times10^{15}\ \mathrm{rad\,s^{-1}}}. \] Because \(\omega=2\pi c/\lambda\), \[ \lambda_{\mathrm p}=\frac{2\pi c}{\omega_{\mathrm p}} =2.39\times10^{-7}\ \mathrm m. \] Thus \[ \boxed{\lambda_{\mathrm p}=239\ \mathrm{nm}}. \]

2.8a

In the stated collision model, each electron loses its entire collective kinetic energy in a collision. Its time-averaged energy loss per collision is \[ \frac12m_e\langle v^2\rangle. \] There are, on average, \(N/\tau\) collisions per second among all electrons. Random collision times allow us to use the cycle-averaged kinetic energy, giving \[ \boxed{ P_{\mathrm{heat}} =\frac{Nm_e}{2\tau}\langle v^2\rangle =\frac{m_0}{2\tau}\langle v^2\rangle}. \]

From the current–velocity relation, \[ \boxed{\langle I^2\rangle =(ne\pi R^2)^2\langle v^2\rangle}. \]

2.8b

The equivalent resistance is defined by \[ P_{\mathrm{heat}}=R_{\mathrm{heat}}\langle I^2\rangle. \] Consequently, \[ R_{\mathrm{heat}} =\frac{m_0}{2\tau(ne\pi R^2)^2} =\frac{L}{2\tau}. \] Thus \[ \boxed{ R_{\mathrm{heat}} =\frac{2m_e}{3\pi ne^2R\tau} =2.46\ \Omega}. \]

The factor \(1/2\) follows from the kinetic energy \(\frac12m_ev^2\) lost per collision in the specific model given in the problem.

2.9

For harmonic motion of amplitude \(x_0\), \[ x_{\mathrm p}(t)=x_0\cos(\omega_{\mathrm p}t+\phi), \] and hence \[ \langle v^2\rangle=\frac12\omega_{\mathrm p}^2x_0^2. \] Therefore \[ \langle I^2\rangle =\frac12(ne\pi R^2)^2\omega_{\mathrm p}^2x_0^2. \]

Define the scattering resistance by \[ P_{\mathrm{scat}}=R_{\mathrm{scat}}\langle I^2\rangle. \] Using the supplied radiation formula, \[ R_{\mathrm{scat}} = \frac{Q^2x_0^2\omega_{\mathrm p}^4} {12\pi\varepsilon_0c^3} \frac{2}{(ne\pi R^2)^2\omega_{\mathrm p}^2x_0^2}. \] Since \[ Q=-\frac43\pi R^3ne, \qquad \frac{Q^2}{(ne\pi R^2)^2}=\frac{16R^2}{9}, \] we obtain \[ \boxed{ R_{\mathrm{scat}} =\frac{8R^2\omega_{\mathrm p}^2} {27\pi\varepsilon_0c^3} =2.45\ \Omega}. \] Equivalently, \[ R_{\mathrm{scat}} =\frac{8ne^2R^2}{81\pi\varepsilon_0^2m_ec^3}. \]

2.10a

Both heating and scattering remove energy from the same oscillating current. Their equivalent resistances therefore add in the series circuit: \[ R_\Sigma=R_{\mathrm{heat}}+R_{\mathrm{scat}}. \]

At \(\omega=\omega_{\mathrm p}=1/\sqrt{LC}\), the inductive and capacitive reactances cancel: \[ \omega_{\mathrm p}L-\frac1{\omega_{\mathrm p}C}=0. \] Thus the current is in phase with the driving voltage and has amplitude \[ I_0=\frac{V_0}{R_\Sigma} =\frac{4RE_0}{3R_\Sigma}. \] For a sinusoidal current, \[ \langle I^2\rangle=\frac{I_0^2}{2} =\frac{8R^2E_0^2}{9R_\Sigma^2}. \] The two average powers are therefore \[ \boxed{ P_{\mathrm{heat}}= \frac{8R^2E_0^2}{9} \frac{R_{\mathrm{heat}}} {(R_{\mathrm{heat}}+R_{\mathrm{scat}})^2}} \] and \[ \boxed{ P_{\mathrm{scat}}= \frac{8R^2E_0^2}{9} \frac{R_{\mathrm{scat}}} {(R_{\mathrm{heat}}+R_{\mathrm{scat}})^2}}. \]

2.10b

From the specified incident intensity, \[ S=\frac12c\varepsilon_0E_0^2, \] so \[ E_0=\sqrt{\frac{2S}{c\varepsilon_0}} =\sqrt{\frac{2(1.00\times10^6)} {(2.998\times10^8)(8.854\times10^{-12})}}. \] Thus \[ \boxed{E_0=2.74\times10^4\ \mathrm{V\,m^{-1}}}. \]

Using unrounded intermediate values, \[ V_0=3.66\times10^{-4}\ \mathrm V, \qquad R_\Sigma=4.910\ \Omega, \qquad I_0=7.45\times10^{-5}\ \mathrm A. \] Consequently, \[ \boxed{P_{\mathrm{heat}}=6.82\times10^{-9}\ \mathrm W}, \qquad \boxed{P_{\mathrm{scat}}=6.82\times10^{-9}\ \mathrm W}. \] Their near equality follows from the near equality of the two resistances.

The displacement amplitude can also be checked: \[ x_0=\frac{I_0}{ne\pi R^2\omega_{\mathrm p}} =3.21\times10^{-15}\ \mathrm m, \] so \(x_0/R=3.21\times10^{-7}\ll1\). Also, \(\omega_{\mathrm p}\tau=41.3\), consistent with the assumed separation of oscillation and collision timescales.

2.11a

Apply the per-particle heating power from 2.10 throughout the illuminated volume. This uses uniform illumination and does not introduce an additional model for attenuation or multiple scattering.

The vessel volume is \[ V_{\mathrm{vessel}}=h^2a =(0.10)^2(0.010) =1.0\times10^{-4}\ \mathrm{m^3}. \] It contains \[ N_{\mathrm{np}}=n_{\mathrm{np}}h^2a =7.3\times10^{11} \] nanoparticles. All their Joule heating is assumed to produce steam, so \[ P_{\mathrm{st}}=N_{\mathrm{np}}P_{\mathrm{heat}} =4.98\times10^3\ \mathrm W. \]

Each kilogram that becomes steam must first be heated from \(20^\circ\mathrm C\) to \(100^\circ\mathrm C\), then vaporized, and finally heated to \(110^\circ\mathrm C\). The required energy per unit mass is \[ q_{\mathrm{st}} =c_{\mathrm{wa}}(T_{100}-T_{\mathrm{wa}}) +L_{\mathrm{wa}} +c_{\mathrm{st}}(T_{\mathrm{st}}-T_{100}). \] Numerically, \[ \begin{aligned} q_{\mathrm{st}} &=(4.181\times10^3)(80) +2.260\times10^6 +(2.080\times10^3)(10)\\ &=2.61528\times10^6\ \mathrm{J\,kg^{-1}}. \end{aligned} \] The bulk water remaining at \(20^\circ\mathrm C\) does not eliminate the heating required for the portion that evaporates.

Energy conservation gives \[ \mu_{\mathrm{st}}q_{\mathrm{st}}=P_{\mathrm{st}}. \] Thus \[ \mu_{\mathrm{st}} =\frac{n_{\mathrm{np}}h^2aP_{\mathrm{heat}}} {c_{\mathrm{wa}}(T_{100}-T_{\mathrm{wa}}) +L_{\mathrm{wa}} +c_{\mathrm{st}}(T_{\mathrm{st}}-T_{100})} =1.904\times10^{-3}\ \mathrm{kg\,s^{-1}}. \] To the precision of the vessel and concentration data, \[ \boxed{\mu_{\mathrm{st}}\simeq1.9\ \mathrm{g\,s^{-1}}}. \]

2.11b

The beam enters normally through the face of area \(h^2\), so \[ P_{\mathrm{tot}}=Sh^2 =(1.00\times10^6)(0.10)^2 =1.00\times10^4\ \mathrm W. \] Therefore \[ \eta=\frac{P_{\mathrm{st}}}{P_{\mathrm{tot}}} =\frac{n_{\mathrm{np}}h^2aP_{\mathrm{heat}}}{Sh^2} =\frac{n_{\mathrm{np}}aP_{\mathrm{heat}}}{S}. \] Numerically, \[ \eta=0.4979, \] or, with sensible rounding, \[ \boxed{\eta\simeq0.50=50\%}. \]

The solution was checked against the official solution.