Physics

The block moves from one endpoint to the other in half a period, so $T/2 = 0.15$ s and $T = 0.30$ s. The endpoints are 40 cm apart, which is twice the amplitude.

(a) Frequency.

$f = \frac{1}{T} = \frac{1}{0.30} = 3.33\ \text{Hz}$

(b) Amplitude.

$A = \frac{0.40}{2} = 0.20\ \text{m} = 20\ \text{cm}$

(c) Force constant. The angular frequency is $\omega = 2\pi f = 2\pi/0.30 = 20.94$ rad/s. Using $\omega^2 = k/m$:

$k = m\omega^2 = 0.250\times(20.94)^2 = 0.250\times438.6 \approx 110\ \text{N/m}$

Given: $m=250\ \text{g}=0.25$ kg; time between the two endpoints $=0.15$ s; distance between endpoints $=40$ cm.

(a) Frequency. Moving from one endpoint to the other takes half a period, so $T/2=0.15$ s and

$T=0.30\ \text{s},\qquad f=\frac1T=\frac1{0.30}\approx3.33\ \text{Hz}.$

(b) Amplitude. The endpoints are $2A$ apart:

$A=\frac{40}{2}=20\ \text{cm}=0.20\ \text{m}.$

(c) Force constant. The angular frequency is

$\omega=2\pi f=\frac{2\pi}{0.30}=20.94\ \text{rad/s}.$

Since $\omega=\sqrt{k/m}$,

$k=m\omega^2=0.25\times(20.94)^2=0.25\times438.6\approx110\ \text{N/m}.$

$y = (CB + \overline{C}A)\,(\overline{BA})$

Circuit (AND, OR, NOT gates):

1. A NOT gate on $C$ gives $\overline{C}$.
2. AND gate 1: inputs $C$, $B$ gives $CB$.
3. AND gate 2: inputs $\overline{C}$, $A$ gives $\overline{C}A$.
4. OR gate: inputs $CB$ and $\overline{C}A$ gives $P = CB + \overline{C}A$.
5. AND gate 3: inputs $B$, $A$ gives $BA$; a NOT gate on it gives $\overline{BA}$.
6. Final AND gate: inputs $P$ and $\overline{BA}$ gives $y$.

This uses 3 AND, 1 OR and 2 NOT gates for the first part, plus one more AND for the last stage (four AND gates in total).

Truth table:

Hence $y = 1$ only for $(A,B,C) = (0,1,1)$ and $(1,0,0)$, i.e. $y = \overline{A}BC + A\overline{B}\,\overline{C}$.

$y=(A'B'+B'A)\,\big((A+B)'+C\big)$

Circuit. The gates needed are:

• NOT gates on $A$ and $B$ to give $A'$ and $B'$.
• Two AND gates, $A'\cdot B'$ and $B'\cdot A$, feeding an OR gate to give $X = A'B' + B'A$.
• An OR gate on $A,B$ followed by a NOT gate to give $(A+B)'$, then an OR gate with $C$ to give $Z=(A+B)'+C$.
• A final AND gate combining $X$ and $Z$ to give $y = X\cdot Z$.

Truth table. Here $X = A'B'+AB'$ and $Z=(A+B)'+C$.

Simplification (check). $A'B'+AB' = B'$ and $(A+B)'=A'B'$, so $y=B'(A'B'+C)=A'B'+B'C$. This is 1 for $ABC=000, 001, 101$, which agrees with the table.

The block takes half a period to travel between the endpoints, so $T/2=0.2$ s and $T=0.4$ s. The endpoints are 50 cm apart, which is twice the amplitude.

Frequency.

$f=\frac1T=\frac1{0.4}=2.5\ \text{Hz}$

Amplitude.

$A=\frac{0.50}{2}=0.25\ \text{m}=25\ \text{cm}$

Force constant. $\omega=2\pi f=5\pi=15.71$ rad/s.

$k=m\omega^2=0.250\times(15.71)^2=0.250\times246.7\approx 61.7\ \text{N/m}$

$y=(C+B'A)\,(A+B+D)'$

Simplification. $(A+B+D)'=A'B'D'$. Then

$y=(C+B'A)A'B'D'=C\,A'B'D'+B'A\cdot A'B'D'$

Since $A\cdot A'=0$, the second term vanishes, so $y=A'B'CD'$.

Circuit.

• NOT gate on $B$ to give $B'$, then an AND gate ($B'\cdot A$).
• An OR gate combining this with $C$ to give $P=C+B'A$.
• A 3-input OR gate $A+B+D$ followed by a NOT gate to give $Q=(A+B+D)'$.
• A final AND gate $y=P\cdot Q$.

Truth table.

The output is 1 only for $ABCD=0010$, which agrees with $y=A'B'CD'$.

Simplification. Using $\overline{CB}=\bar C+\bar B$ and $\overline{B+C}=\bar B\,\bar C$:

$y=\big(AB+A(\bar B+\bar C)\big)\cdot\bar B\,\bar C$

Expand: $AB\cdot\bar B\bar C=0$, and $A(\bar B+\bar C)\bar B\bar C=A\bar B\bar C$. Therefore

$y=A\,\bar B\,\bar C$

Circuit (AND, OR, NOT gates).

1. $G_1$: AND of $A$ and $B$ gives $AB$.
2. $G_2$: AND of $C$ and $B$ gives $CB$. $G_3$: NOT of $G_2$ gives $\overline{CB}$.
3. $G_4$: AND of $\overline{CB}$ and $A$ gives $\overline{CB}\cdot A$.
4. $G_5$: OR of $G_1$ and $G_4$ gives $AB+\overline{CB}A$.
5. $G_6$: OR of $B$ and $C$, followed by $G_7$: NOT, gives $\overline{B+C}$.
6. $G_8$: AND of $G_5$ and $G_7$ gives the output $y$.

Truth table.

The output is 1 only for $A=1,\ B=0,\ C=0$, which agrees with $y=A\bar B\bar C$.

(a) Number of states for $n=3$. There are $2n^2=2(9)=18$ states.

(b) Distribution among subshells. For $n=3$, $l=0,1,2$. For each $l$, $m_l=-l,\dots,+l$ and $m_s=\pm\frac12$, giving $2(2l+1)$ states.

Total $=2+6+10=18$.

Labelling states as $(n,l,m_l,m_s)$:

• 3s: $(3,0,0,+\tfrac12)$, $(3,0,0,-\tfrac12)$
• 3p: $(3,1,m_l,\pm\tfrac12)$ with $m_l=-1,0,+1$ (six states)
• 3d: $(3,2,m_l,\pm\tfrac12)$ with $m_l=-2,-1,0,1,2$ (ten states)

(c) General result. For a given $n$, $l$ runs from $0$ to $n-1$, and each $l$ has $2(2l+1)$ states, so

$N=\sum_{l=0}^{n-1}2(2l+1)=2\left[2\frac{(n-1)n}{2}+n\right]=2\left[n^2-n+n\right]=2n^2.$

Differential equation

A mass $m$ attached to a spring of force constant $k$ on a frictionless surface is displaced by $x$ from equilibrium. Hooke's law gives the restoring force $F=-kx$. Newton's second law, $F=m\,d^2x/dt^2$, gives

$m\frac{d^2x}{dt^2}=-kx\ \Rightarrow\ \frac{d^2x}{dt^2}+\omega^2x=0,\qquad \omega^2=\frac km.$

General solution

Try $x=e^{\lambda t}$: $\lambda^2+\omega^2=0$, so $\lambda=\pm i\omega$ and

$x(t)=C_1\cos\omega t+C_2\sin\omega t=A\cos(\omega t+\phi),$

where $A=\sqrt{C_1^2+C_2^2}$ is the amplitude and $\phi$ the phase constant, fixed by initial conditions.

Period. $x(t+T)=x(t)$ requires $\omega T=2\pi$:

$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac mk},\qquad f=\frac1T=\frac1{2\pi}\sqrt{\frac km}.$

Velocity

$v=\frac{dx}{dt}=-A\omega\sin(\omega t+\phi)$

with maximum $v_{\max}=A\omega$ (at the equilibrium position), and in terms of position $v=\pm\omega\sqrt{A^2-x^2}$.

Acceleration

$a=\frac{d^2x}{dt^2}=-A\omega^2\cos(\omega t+\phi)=-\omega^2x$

with maximum $a_{\max}=A\omega^2$ (at the extreme positions). The velocity leads the displacement by $\pi/2$, and the acceleration is opposite in phase to the displacement.

Energy. $E=\tfrac12mv^2+\tfrac12kx^2=\tfrac12kA^2$, which is constant.

Construction

A bipolar junction transistor (BJT) is a three-layer, two-junction device in which both electrons and holes take part in conduction. It is either n-p-n or p-n-p. The three regions are:

• Emitter (E): heavily doped, emits majority carriers.
• Base (B): very thin ($\sim1\ \mu$m) and lightly doped, so few carriers recombine in it.
• Collector (C): moderately doped, larger in area, collects the carriers.

The emitter-base junction is forward biased and the collector-base junction is reverse biased in normal (active) operation.

Working (n-p-n)

1. The forward-biased emitter-base junction injects many electrons from the emitter into the base (and a few holes from the base to the emitter).
2. The base is thin and lightly doped, so only a small fraction ($\sim1\%$) of the electrons recombine with holes, giving a small base current $I_B$. The rest diffuse across the base.
3. The reverse-biased collector junction has a strong field that sweeps these electrons into the collector, producing the collector current $I_C$.

Current relations:

$I_E=I_B+I_C,\qquad \alpha=\frac{I_C}{I_E}\ (\approx0.95\text{-}0.99),\qquad \beta=\frac{I_C}{I_B}=\frac{\alpha}{1-\alpha}\ (\text{tens to hundreds}).$

A small change in base current causes a large change in collector current, so the transistor amplifies current. In a common-emitter amplifier a small input signal across the base-emitter gives a larger voltage change across the collector load $R_L$. The device also works as a switch, being cut off ($I_B=0$) or saturated (large $I_B$). A p-n-p transistor works the same way with holes as the main carriers and all polarities reversed.

The probability density is

$|\psi|^2 = \frac{2}{a}\sin^2\!\left(\frac{n\pi x}{a}\right)$

and the probability of finding the particle in a small interval $dx$ about $x$ is $|\psi|^2\,dx$. At $x = a/4$:

$|\psi|^2 = \frac{2}{a}\sin^2\!\left(\frac{n\pi}{4}\right)$

n = 1: $\sin^2(\pi/4) = \tfrac12$, so $|\psi|^2 = \dfrac{1}{a}$.

n = 2: $\sin^2(\pi/2) = 1$, so $|\psi|^2 = \dfrac{2}{a}$.

n = 3: $\sin^2(3\pi/4) = \tfrac12$, so $|\psi|^2 = \dfrac{1}{a}$.

So the probability per unit length at $x=a/4$ is $1/a,\ 2/a,\ 1/a$ for $n=1,2,3$, and the probability in a narrow width $\Delta x$ is $|\psi|^2\Delta x$ (for example $\Delta x/a$ for $n=1$). The $n=2$ state has an antinode at $a/4$, hence the largest value; for $n=1$ and $n=3$ the value is half of that.

Use Bragg's law for the first order ($n=1$):

$2d\sin\theta = n\lambda$

with $d = 0.732$ Å and $\lambda = 0.1$ Å:

$\sin\theta = \frac{\lambda}{2d} = \frac{0.1}{2(0.732)} = 0.0683$

$\theta = \sin^{-1}(0.0683) \approx 3.9^\circ$

The neutron beam must make an angle of about 3.9° with the atomic planes (glancing angle) to reflect neutrons of wavelength 0.1 Å.

Effective mass

In a crystal, an electron moves in the periodic potential of the lattice as well as under an applied force. It is convenient to describe this by treating the electron as a free particle with a modified mass, the effective mass $m^*$.

For an electron described by a wave packet, with energy $E(k)$ and group velocity $v_g=\dfrac1\hbar\dfrac{dE}{dk}$, an applied force $F$ does work $F\,v_g\,dt=dE$, which gives $\hbar\,dk/dt=F$. The acceleration is

$a=\frac{dv_g}{dt}=\frac1\hbar\frac{d^2E}{dk^2}\frac{dk}{dt}=\frac{1}{\hbar^2}\frac{d^2E}{dk^2}F.$

Comparing with $F=m^*a$:

$m^*=\frac{\hbar^2}{d^2E/dk^2}.$

So the effective mass is inversely proportional to the curvature of the $E$-$k$ band.

Electrons

Near the bottom of the conduction band the curvature is positive, so $m_e^*>0$ and the electron accelerates in the direction of the force. A narrow band with weak curvature gives a large $m^*$; a sharply curved band gives a small $m^*$.

Holes

Near the top of the valence band $d^2E/dk^2<0$, so an electron there would have a negative effective mass. The absence of an electron (a hole) is treated as a particle of positive charge $+e$ and positive effective mass $m_h^*=|m^*|$, which moves in the direction of the applied electric field.

Significance. $m^*$ determines mobility $\mu=e\tau/m^*$, the density of states, and the carrier concentration, so it controls conductivity. For example in Si, $m_e^*\approx0.26\,m_0$ and $m_h^*\approx0.39\,m_0$ (conductivity values), so electrons are more mobile than holes.

Given: $I=50$ A, thickness along $B$: $t=0.5$ cm $=5\times10^{-3}$ m, width $w=2$ cm, $B=1.5$ T, $n=8.4\times10^{28}\ \text{m}^{-3}$.

The field is perpendicular to the slab, so it is along the thickness $t$. The Hall field is set by $E_H=v_dB$, and the Hall voltage across the width is $V_H=E_Hw=v_dBw$. With $v_d=\dfrac{I}{newt}$,

$V_H=\frac{IB}{net}.$

$V_H=\frac{50\times1.5}{(8.4\times10^{28})(1.6\times10^{-19})(5\times10^{-3})}=\frac{75}{6.72\times10^{7}}$

$V_H\approx1.12\times10^{-6}\ \text{V}\approx1.1\ \mu\text{V}.$

Given: $\Delta x=\lambda=\dfrac hp$, where $p=mv$ is the particle's momentum.

By the Heisenberg uncertainty principle,

$\Delta x\,\Delta p\ge\frac\hbar2=\frac{h}{4\pi}.$

Taking the minimum,

$\Delta p=\frac{h}{4\pi\Delta x}=\frac{h}{4\pi\lambda}=\frac{p}{4\pi}.$

Since $\Delta p=m\Delta v$ and $p=mv$,

$\Delta v=\frac{v}{4\pi}.$

So the uncertainty in velocity is $1/4\pi\approx8\%$ of the particle's own velocity $v$.

In terms of the de Broglie wave velocity. The wave (phase) velocity is $u=\dfrac{\omega}{k}=\dfrac{E}{p}=\dfrac{\tfrac12mv^2}{mv}=\dfrac v2$ for a non-relativistic particle (in general $u=c^2/v$). Hence $v=2u$ and

$\Delta v=\frac{2u}{4\pi}=\frac{u}{2\pi}.$

An FCC unit cell contains $4$ atoms (8 corners $\times\frac18$ + 6 faces $\times\frac12$), and with a one-atom basis there are $4$ copper atoms per cube.

Mass of the cell: $\dfrac{4M}{N_A}$. Density: $\rho=\dfrac{4M}{N_Aa^3}$, so

$a^3=\frac{4M}{\rho N_A}=\frac{4\times63.5}{8.96\times6.022\times10^{23}}=\frac{254}{5.396\times10^{24}}=4.707\times10^{-23}\ \text{cm}^3$

$a=\left(47.07\times10^{-24}\right)^{1/3}=3.61\times10^{-8}\ \text{cm}$

$a\approx3.61\ \text{\AA}=0.361\ \text{nm}.$

Magnetic dipole moment

A current loop of area $A$ carrying current $I$ has a magnetic dipole moment

$\vec\mu=IA\,\hat n,$

with direction given by the right-hand rule. In a uniform field $\vec B$ it experiences torque and has potential energy

$\vec\tau=\vec\mu\times\vec B,\qquad U=-\vec\mu\cdot\vec B.$

Atomic origin

Orbital moment. An electron of charge $e$ in an orbit has current $I=e/T=ev/2\pi r$, so $\mu_l=I\pi r^2=\dfrac{e}{2m}L$, and as vectors $\vec\mu_l=-\dfrac{e}{2m}\vec L$. With $L=\sqrt{l(l+1)}\,\hbar$, the unit is the Bohr magneton

$\mu_B=\frac{e\hbar}{2m}=9.27\times10^{-24}\ \text{J/T}.$

The component along $B$ is $\mu_{l,z}=-m_l\mu_B$.

Spin moment. The electron has intrinsic spin with $\vec\mu_s=-g_s\dfrac{e}{2m}\vec S$, $g_s\approx2$, $\mu_{s,z}=\mp\mu_B$.

The total atomic moment comes from combining $\vec L$ and $\vec S$ (the Landé $g$-factor).

Effects on atoms

• In a magnetic field, the energy of a level shifts by $\Delta E=m_jg\mu_BB$, so levels split. This gives the Zeeman effect: spectral lines split into several components.
• In a non-uniform field, atoms experience a force $F_z=\mu_z\,dB/dz$, so a beam splits according to $m$ values. The Stern-Gerlach experiment (silver atoms split into two beams) shows space quantisation and electron spin.
• Atoms with filled shells have zero net moment (diamagnetic); atoms with unpaired electrons have a permanent moment (paramagnetic).

Effects on molecules and materials

• Molecules with a net moment (such as $\text{O}_2$) align partially in a field, producing paramagnetism with $M=\chi H$, $\chi\propto1/T$ (Curie law); thermal motion opposes alignment.
• Molecules with no net moment develop an induced moment opposing the field (diamagnetism).
• In solids, interactions between neighbouring moments lead to ferromagnetism (parallel alignment in domains), antiferromagnetism, and ferrimagnetism.
• Nuclear magnetic moments in molecules are the basis of NMR spectroscopy and MRI.

Given: $q_1=2.5\times10^{-10}$ C, $q_2=5\times10^{-10}$ C, separation $1$ m. The electron starts at the midpoint ($0.5$ m from each) and moves along the line to the point $0.1$ m from $q_2$ (so $0.9$ m from $q_1$). Both charges are positive, so the electron is attracted toward the larger charge $q_2$ (net force at the midpoint points toward $q_2$).

Potentials ($k=8.99\times10^9\ \text{N m}^2/\text{C}^2$):

$V_i=k\left(\frac{q_1}{0.5}+\frac{q_2}{0.5}\right)=8.99\times10^9\left(5\times10^{-10}+10^{-9}\right)=13.5\ \text{V}$

$V_f=k\left(\frac{q_1}{0.9}+\frac{q_2}{0.1}\right)=8.99\times10^9\left(2.78\times10^{-10}+5\times10^{-9}\right)=47.4\ \text{V}$

Energy conservation. For the electron (charge $-e$), $\Delta K=-\Delta U=e(V_f-V_i)$:

$\tfrac12mv^2=(1.602\times10^{-19})(47.4-13.5)=(1.602\times10^{-19})(33.96)=5.44\times10^{-18}\ \text{J}$

$v=\sqrt{\frac{2(5.44\times10^{-18})}{9.11\times10^{-31}}}=\sqrt{1.195\times10^{13}}$

$v\approx3.5\times10^{6}\ \text{m/s}$

(about $3.46\times10^6$ m/s, directed toward $q_2$).

Magnetic dipole moment

A current loop of area $A$ carrying current $I$ behaves as a magnetic dipole with moment

$\vec{\mu} = I\vec{A},$

with direction given by the right-hand rule (perpendicular to the loop). In a magnetic field $\vec B$ it experiences a torque and has a potential energy:

$\vec\tau = \vec\mu\times\vec B,\qquad U = -\vec\mu\cdot\vec B.$

Atomic origin: An electron in orbit of radius $r$ with speed $v$ is a current loop, $I = ev/2\pi r$, so the orbital moment is

$\mu_L = IA = \frac{e}{2m}L,\qquad \vec\mu_L = -\frac{e}{2m}\vec L.$

The electron also has a spin moment $\mu_s = -g_s\dfrac{e}{2m}S$ with $g_s\approx 2$. Their natural unit is the Bohr magneton, $\mu_B = \dfrac{e\hbar}{2m} = 9.27\times10^{-24}$ J/T.

Effect on atoms:

• In an external field the atomic moment precesses about $\vec B$ (Larmor precession) and its orientation is quantised (space quantisation), so an energy $U = m_j g_j\mu_B B$ is added to each level.
• Spectral lines split into several components: the Zeeman effect.
• In a non-uniform field a beam of atoms is deflected according to its moment (the Stern-Gerlach experiment), showing that only discrete orientations occur.
• Atoms with a net moment are paramagnetic; atoms in which all moments cancel are diamagnetic (an induced, opposing moment).

Effect on molecules:

• The moment of a molecule is the vector sum of its electronic orbital, spin and nuclear contributions. Most molecules have paired electrons and zero net moment (diamagnetic), while molecules such as $\text{O}_2$ (two unpaired electrons) and NO are paramagnetic.
• In a field, permanent moments tend to align with $\vec B$ against thermal agitation, giving a magnetisation described by Curie's law, $M\propto B/T$.
• Moments of the nuclei are responsible for nuclear magnetic resonance of molecules.

Integrated circuits (ICs) are fabricated on a single-crystal silicon wafer by the planar process, in which many identical circuits are produced at once and each step is repeated with different masks.

1. Epitaxial growth: A thin, lightly doped single-crystal layer (a few µm thick) is grown on the polished p-type substrate by chemical vapour deposition, for example from $\text{SiCl}_4$ or $\text{SiH}_4$ with hydrogen at about 1200 °C. The layer continues the crystal structure of the wafer, and its doping (for example n-type, by adding $\text{PH}_3$) is controlled independently of the substrate. Devices are formed in this layer.

2. Oxidation: The wafer is heated in oxygen or steam at 1000-1200 °C so that an insulating $\text{SiO}_2$ layer grows on the surface. The oxide masks against dopant diffusion, protects the junctions, and serves as the insulating layer under the metal and as the gate dielectric.

3. Oxide removal and pattern definition (photolithography): The oxide is coated with photoresist and exposed to UV light through a mask. After development, the oxide is etched away (HF) only where the resist has been removed, opening windows to the silicon. The resist is then stripped. A different mask is used for each stage (isolation, base, emitter, contacts, and so on).

4. Doping (impurities in Si): Acceptors (boron) or donors (phosphorus, arsenic) are introduced through the windows by thermal diffusion at 900-1200 °C or by ion implantation followed by annealing. This forms the p and n regions, giving the junctions of transistors, diodes and resistors. The oxide stops the dopants from entering elsewhere. Isolation of components is obtained by reverse-biased p-n junctions or by oxide.

5. Interconnection of components: Contact windows are opened in the oxide and a metal film (aluminium, now often copper) is deposited by evaporation or sputtering. A final lithography and etch step leaves only the desired conductor pattern that joins the components into the circuit. Multi-level interconnections are separated by insulating layers. The wafer is finally tested, scribed into chips, mounted and bonded in a package, and a passivation layer protects the surface.

Electron concentration: Aluminium is trivalent, so each atom gives 3 free electrons.

$n = 3\,\frac{\rho N_A}{M} = 3\,\frac{(2.70)(6.022\times10^{23})}{26.98} = 1.81\times10^{23}\ \text{cm}^{-3} = 1.81\times10^{29}\ \text{m}^{-3}$

Fermi energy:

$E_F = \frac{\hbar^2}{2m}\left(3\pi^2n\right)^{2/3}$

$3\pi^2 n = 29.61\times1.81\times10^{29} = 5.35\times10^{30}\ \text{m}^{-3}$

$(5.35\times10^{30})^{2/3} = 3.06\times10^{20}\ \text{m}^{-2}$

$\frac{\hbar^2}{2m} = \frac{(1.055\times10^{-34})^2}{2(9.11\times10^{-31})} = 6.11\times10^{-39}\ \text{J m}^2$

$E_F = (6.11\times10^{-39})(3.06\times10^{20}) = 1.87\times10^{-18}\ \text{J} = 11.7\ \text{eV}$

Effective mass: With $m^*$ replacing $m$, $E_F\propto 1/m^*$. If the experimental value is 12 eV,

$\frac{m^*}{m} = \frac{E_F(\text{calc})}{E_F(\text{exp})} = \frac{11.7}{12} \approx 0.97$

$m^* \approx 0.97\,m \approx 8.9\times10^{-31}\ \text{kg}$

The effective mass is about 0.97 times the free-electron mass.

Meaning of fabrication of integrated circuits

An integrated circuit (IC) is a complete electronic circuit (transistors, diodes, resistors, capacitors and their interconnections) formed together on a single small chip of semiconductor, usually silicon. Fabrication means the sequence of chemical and physical steps by which these components are built into and onto a silicon wafer, hundreds of identical chips at a time, using repeated cycles of film growth, patterning and doping (the planar process).

Processes

1. Epitaxial growth: A thin single-crystal silicon layer with controlled doping is grown on the polished wafer by chemical vapour deposition (for example from $\text{SiCl}_4$ and $\text{H}_2$ at about 1200 °C). It is of the same crystal orientation as the substrate, and the components are built into it. A buried heavily doped layer under it reduces the series resistance of transistors.

2. Oxidation: Heating the wafer in oxygen or steam at 1000-1200 °C grows a layer of $\text{SiO}_2$. This oxide insulates, masks against diffusion of impurities, passivates the surface and forms the dielectric of MOS gates and capacitors.

3. Oxide removal and pattern definition (photolithography): The oxide is coated with a photoresist, exposed to ultraviolet light through a mask carrying the circuit pattern, and developed. The exposed oxide is etched away with HF, producing windows in the oxide, and the resist is then removed. Each stage has its own mask, aligned exactly with the previous pattern.

4. Doping: Donor (P, As) or acceptor (B) atoms are introduced through the oxide windows by high-temperature diffusion (900-1200 °C) or by ion implantation followed by annealing. This creates the n- and p-type regions that form the junctions (base, emitter, resistor regions, source and drain). Oxide-protected regions remain undoped.

5. Interconnection of components: Contact windows are opened in the oxide, and a metal layer (aluminium or copper) is deposited by evaporation or sputtering over the wafer. Another lithography and etching step leaves metal strips that wire the components together; several metal layers separated by insulating oxide can be built up. A passivation layer protects the chip, which is then tested, cut from the wafer, bonded and packaged.

Effect of an external magnetic field on current-carrying loops

A current loop placed in a uniform magnetic field experiences no net force, because the forces on opposite sides are equal and opposite:

$\vec F_\text{net} = \oint I\,d\vec L\times\vec B = I\left(\oint d\vec L\right)\times\vec B = 0$

However, the forces on opposite sides can form a couple, so the loop experiences a torque which tends to turn it so that its plane is perpendicular to the field (the loop's magnetic moment $\vec\mu = I\vec A$ lines up with $\vec B$). This is the principle of the galvanometer and the electric motor.

Torque on a rectangular loop on a pivot

Let the loop have sides $a$ (vertical, length $a$) and $b$ (horizontal) and carry current $I$. It is pivoted on a vertical axis through its centre, in a uniform field $\vec B$ perpendicular to the axis. Let $\theta$ be the angle between the normal to the loop and $\vec B$.

• The two sides of length $b$ (parallel to the pivot axis' perpendicular direction) have forces along the axis, equal and opposite and collinear, so they produce no torque.
• The two sides of length $a$ are perpendicular to $\vec B$ and each experiences a force of magnitude

$F = IaB$

in opposite directions, forming a couple. The perpendicular distance between their lines of action is $b\sin\theta$. Hence

$\tau = F\,b\sin\theta = IaB\,b\sin\theta = I(ab)B\sin\theta = IAB\sin\theta$

In vector form:

$\vec\tau = I\vec A\times\vec B = \vec\mu\times\vec B$

The torque is maximum, $IAB$, when the plane of the loop is parallel to $\vec B$ ($\theta=90^\circ$), and zero when the plane is perpendicular to $\vec B$ ($\theta = 0$), the stable equilibrium.

Increasing the torque

For a coil of $N$ turns,

$\tau = NIAB\sin\theta$

so the torque can be increased by using more turns $N$ (an alternative to increasing $I$, $A$ or $B$), by increasing the area, or by placing the coil on a soft-iron core, which increases the field in which it sits. In a moving-coil galvanometer a radial field is used so that the plane of the coil is always parallel to $\vec B$ and $\tau = NIAB$ for all positions.

The proton gains kinetic energy equal to the potential energy lost in moving through the potential difference. Starting from rest:

$qV = \tfrac12 mv^2 \quad\Rightarrow\quad v=\sqrt{\frac{2qV}{m}}$

$v=\sqrt{\frac{2\,(1.6\times10^{-19})(100)}{1.67\times10^{-27}}}=\sqrt{\frac{3.2\times10^{-17}}{1.67\times10^{-27}}}=\sqrt{1.916\times10^{10}}$

$v \approx 1.38\times10^{5}\ \text{m/s}$

The proton's kinetic energy is $100$ eV, and $v \ll c$, so the non-relativistic formula is valid.

(a) Fermi energy. Aluminium is trivalent, so each atom gives 3 free electrons.

$n = 3\,\frac{\rho N_A}{M} = 3\times\frac{2.70\times6.022\times10^{23}}{26.98} = 1.808\times10^{23}\ \text{cm}^{-3} = 1.808\times10^{29}\ \text{m}^{-3}$

$E_F=\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$

$3\pi^2 n = 29.61\times1.808\times10^{29} = 5.353\times10^{30}\ \text{m}^{-3}$

$(3\pi^2 n)^{2/3} = (1.749\times10^{10})^2 = 3.06\times10^{20}\ \text{m}^{-2}$

$\frac{\hbar^2}{2m}=\frac{(1.055\times10^{-34})^2}{2(9.109\times10^{-31})}=6.10\times10^{-39}\ \text{J m}^2$

$E_F = 6.10\times10^{-39}\times3.06\times10^{20}=1.87\times10^{-18}\ \text{J}\approx 11.7\ \text{eV}$

(b) Effective mass. Since $E_F \propto 1/m$, replacing $m$ by $m^*$ gives $E_F^{exp}/E_F^{free} = m/m^*$:

$\frac{m^*}{m}=\frac{11.7}{12}\approx 0.97 \quad\Rightarrow\quad m^*\approx 0.97\,m_e \approx 8.9\times10^{-31}\ \text{kg}$

The particle starts from rest, and the work done by the field equals its gain in kinetic energy:

$qV=\tfrac12mv^2\quad\Rightarrow\quad v=\sqrt{\frac{2qV}{m}}$

Using $q=3.2\times10^{-19}$ C, $V=100$ V and the mass of the alpha particle $m=6.70\times10^{-27}$ kg (the exponent printed in the question is a misprint, since a mass of $10^{-19}$ kg is not that of a particle):

$v=\sqrt{\frac{2(3.2\times10^{-19})(100)}{6.70\times10^{-27}}}=\sqrt{\frac{6.4\times10^{-17}}{6.70\times10^{-27}}}=\sqrt{9.55\times10^{9}}$

$v\approx 9.8\times10^{4}\ \text{m/s}$

The kinetic energy gained is $qV=200$ eV.