Read the following passage and answer the questions that follow :

Diffusion of impurity atoms in silicon is important in silicon integrated circuit processing. The idea of using diffusion techniques to alter the type of conductivity in silicon or germanium was disclosed in a patent by Pfann. Since then various ways of introducing dopants into silicon by diffusion have been studied with the goal of controlling dopant distribution, total dopant concentration, its uniformity reproducibility, and of processing a large number of device wafers in a batch to reduce the manufacturing cost. The diffusion is used to form bases, emitters and resistors in bipolar devices technology, to form source and drain regions and to dope polysilicon in MOS device technology. Dopant atoms that span a wide range of concentrations can be introduced into silicon in many ways.

 William Pfann disclosed the use of diffusion to alter the conductivity type of silicon and germanium in a patent of 1952 — option 2.

Where the date sits in the history of the technology, which is what makes it memorable rather than arbitrary:

Pfann's own major contribution was zone refining, the technique of passing a molten zone repeatedly along an ingot so that impurities are swept to one end. It is what first made semiconductor-grade material pure enough to be useful, and it is the necessary complement to diffusion: one removes unwanted impurities, the other introduces wanted ones under control.

Why diffusion mattered so much. Before it, junctions were made by alloying or by grown-junction methods, in which the junction depth was poorly controlled and set by mechanical or crystal-growth conditions. Diffusion made the depth a function of temperature and time alone:

\(x_{j}\propto\sqrt{Dt}\)

— both quantities that a furnace controls precisely. Thin, reproducible base regions became possible, and with them high-frequency transistors; and because the process works on a whole wafer at once through an oxide mask, it is what made batch manufacture possible. The passage's phrase about "processing a large number of device wafers in a batch to reduce the manufacturing cost" points at exactly this.

The date's significance in one line : it falls between the invention of the transistor and the invention of the integrated circuit, and it is one of the steps that made the second possible.

Hence, the year is 1952.

This is the limited-source or drive-in case, whose profile is Gaussian — and the surface concentration is that Gaussian evaluated at x = 0.

The profile. With a fixed total dose Q_T deposited in a thin layer and then driven in, the solution of the diffusion equation is

\(C(x,t)=\dfrac{Q_{T}}{\sqrt{\pi Dt}}\exp\left(\dfrac{-x^{2}}{4Dt}\right)\)

Setting x = 0 gives the surface value, since the exponential becomes 1:

\(C_{S}=\dfrac{Q_{T}}{\sqrt{\pi Dt}}\)

— option 3. Option 2 is the trap: it is the complete profile \(C(x,t)\) rather than the surface concentration, so it still carries the x-dependent exponential. The question asks specifically for \(C_{S}\).

Two checks that confirm the form.

Dimensions. Q_T is a dose per unit area, in cm^–2. Since Dt has units of cm², \(\sqrt{\pi Dt}\) is a length, and dividing gives cm^–3 — a concentration, as required. Option 4 divides by Dt rather than its square root and yields the wrong units; option 1 equates a volume concentration to an areal dose.

Conservation. Integrating the Gaussian over all depth returns exactly Q_T, which is the defining property of this case: no dopant enters or leaves, it merely spreads.

The physical consequence is important. Because Q_T is fixed, the surface concentration falls as \(1/\sqrt{t}\) while the profile deepens as \(\sqrt{Dt}\). Driving in longer spreads the same dopant further and dilutes the surface — exactly what is wanted when a lightly doped, deep region such as a well or a bipolar base is needed.

Contrast with the other standard case. A constant-source diffusion holds the surface at the solid solubility limit and lets dopant keep entering; its profile is a complementary error function, its surface concentration is fixed, and the dose grows as \(\sqrt{t}\). Real processes use both in sequence — a short constant-source predeposition to meter the dose, then a limited-source drive-in to place it.

Hence, C_S = Q_T/√(πDt).

Sheet resistance is a resistance, so it must be the reciprocal of an integrated conductance — option 3:

\(R_{S}=\dfrac{1}{q\displaystyle\int_{0}^{x_{j}}\mu\,C(x)\,dx}\)

Building it from the conductivity. At any depth the local conductivity is

\(\sigma(x)=q\,\mu\,C(x)\)

Since the layer is non-uniformly doped, the conductances of its thin sub-layers add in parallel, which means integrating the conductivity through the layer:

\(G_{sheet}=\int_{0}^{x_{j}}\sigma(x)\,dx=q\int_{0}^{x_{j}}\mu\,C(x)\,dx\)

Sheet resistance is the reciprocal of that, which is exactly option 3.

Why the reciprocal is the decisive feature. Options 1 and 2 place the integral in the numerator, so a more heavily doped layer would have a higher resistance — the opposite of the truth. Adding dopant adds carriers and must lower the resistance, so the integral has to be in the denominator. That single test disposes of half the options.

Why the limits run only to x_j. The junction depth is where the diffused dopant concentration falls to the background level of the substrate. Beyond it the material is of the opposite type and is isolated by the reverse-biased junction, so it carries none of the layer's current and must be excluded.

Why the temperature factor in options 2 and 4 is out of place. The \(RT/q\) group is a thermal voltage, and it belongs to the Einstein relation between mobility and diffusivity, not to a resistance calculation. Temperature does affect \(\mu\), but it enters through the mobility already inside the integral.

How it is used. Sheet resistance is quoted in "ohms per square", the unit being dimensionless because the resistance of a square patch is independent of its size:

\(R=R_{S}\dfrac{L}{W}\)

so a diffused resistor's value is set purely by counting squares in the layout. Irvin's curves relate \(R_{S}\), \(x_{j}\) and the surface concentration, so measuring the sheet resistance with a four-point probe gives the doping profile without sectioning the wafer.

Hence, R_S is the reciprocal of the integrated conductivity.

Fick's second law is a time derivative on the left and a space derivative twice on the right — option 1:

\(\dfrac{\partial C}{\partial t}=\dfrac{\partial}{\partial x}\left[D\dfrac{\partial C}{\partial x}\right]\)

Where it comes from — two statements combined.

Fick's first law says that flux is driven by a concentration gradient, and flows down it:

\(J=-D\dfrac{\partial C}{\partial x}\)

Continuity says that whatever does not flow out must accumulate:

\(\dfrac{\partial C}{\partial t}=-\dfrac{\partial J}{\partial x}\)

Substituting the first into the second gives the second law directly, and the two minus signs cancel. Seeing this derivation makes the structure inevitable: the left side is a rate of accumulation in time, and the right is the divergence of a flux in space. The two sides must therefore carry different variables, which immediately rules out options 2, 3 and 4 — each of which differentiates twice with respect to the same variable somewhere.

Checking option 1 dimensionally. With C in cm^–3 and D in cm²/s:

\(\left[\dfrac{\partial C}{\partial t}\right]=\dfrac{\text{cm}^{-3}}{\text{s}},\qquad \left[\dfrac{\partial^{2}C}{\partial x^{2}}\right]=\dfrac{\text{cm}^{-3}}{\text{cm}^{2}}\)

and multiplying the second by D's cm²/s reproduces the first exactly. None of the other three balances.

Why D is left inside the derivative. If diffusivity is constant the equation simplifies to \(\partial C/\partial t=D\,\partial^{2}C/\partial x^{2}\), whose solutions are the Gaussian and error-function profiles of the previous questions. But D is not always constant: at concentrations above the intrinsic carrier density it becomes concentration dependent through the charged point defects that carry the dopant, so \(D=D(C)\) and the general form must be kept. That is why heavily doped profiles are box-shaped rather than Gaussian.

The same equation governs heat conduction, with temperature in place of concentration — which is why solutions worked out for one field transfer directly to the other.

Hence, the correct expression is option 1.

 Diffusivity follows the Arrhenius law — option 3:

\(D=D_{0}\exp\left(-\dfrac{E}{RT}\right)\)

Why the exponent must be negative. A dopant atom moves only by surmounting an energy barrier — it must break free of its lattice site or wait for a vacancy. The fraction of atoms with enough thermal energy to do so is the Boltzmann factor \(e^{-E/RT}\), which rises as T rises because the exponent becomes less negative. Option 1's positive exponent would make D fall with temperature, and its argument \(RT/q\) is a voltage rather than a dimensionless ratio in any case.

Why the exponent must be dimensionless. Both E and RT are energies per mole, so their ratio is a pure number, as an exponent must be. Option 2 divides \(D_{0}\), a diffusivity, by an energy, and option 4 divides an energy by a temperature — neither is dimensionless, so both fail before any physics is considered.

How the constants are measured. Taking logarithms turns the law into a straight line:

\(\ln D=\ln D_{0}-\dfrac{E}{R}\left(\dfrac{1}{T}\right)\)

Plotting \(\ln D\) against \(1/T\) gives a slope of \(-E/R\) and an intercept of \(\ln D_{0}\). This Arrhenius plot is exactly how the tabulated values for each dopant are obtained — which is what the question means by "determined experimentally over a range of diffusion temperatures".

How steep the dependence is. With \(E\approx3.5\) eV, a rise from 1000 °C to 1100 °C — less than 8 % in absolute temperature — multiplies D by roughly a factor of ten. This is why diffusion furnaces are controlled to within a degree, and why the junction depth \(x_{j}\propto\sqrt{Dt}\) is set far more sensitively by temperature than by time.

Hence, the correct expression is D = D₀ exp(−E/RT).