91Ó°ÊÓ

If an amount of energy \(Q_{o}^{\prime \prime}\left(\mathrm{J} / \mathrm{m}^{2}\right)\) is released instantaneously, as, for example, from a pulsed laser, and it is absorbed by the surface of a semi-infinite medium, with no attendant losses to the surroundings, the subsequent temperature distribution in the medium is $$ T(x, t)-T_{i}=\frac{Q_{o}^{\prime \prime}}{\rho c(\pi \alpha t)^{1 / 2}} \exp \left(-x^{2} / 4 \alpha t\right) $$ where \(T_{i}\) is the initial, uniform temperature of the medium. Consider an analogous mass transfer process involving deposition of a thin layer of phosphorous (P) on a silicon (Si) wafer at room temperature. If the wafer is placed in a furnace, the diffusion of \(\mathrm{P}\) into Si is significantly enhanced by the high-temperature environment. A Si wafer with \(1-\mu \mathrm{m}\)-thick P film is suddenly placed in a furnace at \(1000^{\circ} \mathrm{C}\), and the resulting distribution of \(P\) is characterized by an expression of the form $$ C_{\mathrm{P}}(x, t)=\frac{M_{\mathrm{P}, o}^{\prime \prime}}{\left(\pi D_{\mathrm{P}-\mathrm{Si}} t\right)^{1 / 2}} \exp \left(-x^{2} / 4 D_{\mathrm{P}-\mathrm{Si}} t\right) $$ where \(M_{\mathrm{P}, o}^{\prime \prime}\) is the molar area density \(\left(\mathrm{kmol} / \mathrm{m}^{2}\right)\) of \(\mathrm{P}\) associated with the film of concentration \(C_{\mathrm{P}}\) and thickness \(d_{o}\). (a) Explain the correspondence between variables in the analogous temperature and concentration distributions. (b) Determine the mole fraction of \(P\) at a depth of \(0.1 \mu \mathrm{m}\) in the Si after \(30 \mathrm{~s}\). The diffusion coefficient is \(D_{\mathrm{P}-\mathrm{Si}}=1.2 \times 10^{-17} \mathrm{~m}^{2} / \mathrm{s}\). The mass densities of \(\mathrm{P}\) and \(\mathrm{Si}\) are 2000 and \(2300 \mathrm{~kg} / \mathrm{m}^{3}\), respectively, and their molecular weights are \(30.97\) and \(28.09 \mathrm{~kg} / \mathrm{kmol}\).

Step by step solution

01

Variable correspondence

In both equations, the terms have similar structures, indicating an analogy between variables. We can list the corresponding variables as follows: 1. \(T(x, t) - T_{i}\) (temperature distribution) \(\longleftrightarrow C_{\mathrm{P}}(x, t)\) (concentration distribution) 2. \(Q_{o}^{\prime \prime}\) (energy released per unit area) \(\longleftrightarrow M_{\mathrm{P}, o}^{\prime \prime}\) (molar area density of \(\mathrm{P}\)) 3. \(\alpha\) (thermal diffusivity) \(\longleftrightarrow D_{\mathrm{P}-\mathrm{Si}}\) (diffusion coefficient) 4. \(t\) (time) remains the same in both equations. 5. \(x\) (depth) remains the same in both equations. The analogous variables are associated with the same physical processes: energy distribution in the heat transfer problem and concentration distribution in the mass transfer problem. #b) Determine the mole fraction of \(P\) at a depth of \(0.1 \mu \mathrm{m}\) in the Si after \(30 \mathrm{~s}\)# First, we can calculate \(C_{\mathrm{P}}(x, t)\) using the given formula.

02

Calculate \(C_{\mathrm{P}}(x, t)\)

By substituting the given values into the formula: $$ C_{\mathrm{P}}(x, t)=\frac{M_{\mathrm{P}, o}^{\prime \prime}}{\left(\pi D_{\mathrm{P}-\mathrm{Si}} t\right)^{1 / 2}} \exp \left(-x^{2} / 4 D_{\mathrm{P}-\mathrm{Si}} t\right) $$ We have: - \(x = 0.1 \times 10^{-6} \mathrm{~m}\) - \(t = 30 \mathrm{~s}\) - \(D_{\mathrm{P}-\mathrm{Si}} = 1.2 \times 10^{-17} \mathrm{~m}^{2} / \mathrm{s}\) But first, we need to find the value of \(M_{\mathrm{P}, o}^{\prime \prime}\).

03

Find \(M_{\mathrm{P}, o}^{\prime \prime}\)

To calculate the molar area density, we have: $$ M_{\mathrm{P}, o}^{\prime \prime} = \frac{\rho_{\mathrm{P}} d_{o}}{M_{\mathrm{P}}} $$ where: - \(\rho_{\mathrm{P}}\) is the mass density of \(P\) (\(2000 \mathrm{~kg} / \mathrm{m}^{3}\)) - \(d_{o}\) is the thickness of \(P\) (\(1 \times 10^{-6} \mathrm{~m}\)) - \(M_{\mathrm{P}}\) is the molecular weight of \(P\) (\(30.97 \mathrm{~kg} / \mathrm{kmol}\)) $$ M_{\mathrm{P}, o}^{\prime \prime} = \frac{2000 \mathrm{~kg} / \mathrm{m}^{3} \times 1 \times 10^{-6} \mathrm{~m}}{30.97 \mathrm{~kg} / \mathrm{kmol}} \approx 6.46 \times 10^{-8} \mathrm{~kmol} / \mathrm{m}^{2} $$ Now we can calculate \(C_{\mathrm{P}}(x, t)\).

04

Compute \(C_{\mathrm{P}}(x, t)\)

Given the values above: $$ C_{\mathrm{P}}(0.1 \mu \mathrm{m}, 30 \mathrm{s})=\frac{6.46 \times 10^{-8} \mathrm{kmol/m}^{2}}{\left(\pi (1.2 \times 10^{-17} \mathrm{m}^{2} / \mathrm{s})(30 \mathrm{s})\right)^{1 / 2}} \exp \left(-(0.1 \times 10^{-6} \mathrm{m})^{2} / 4 (1.2 \times 10^{-17} \mathrm{m}^{2} / \mathrm{s})(30 \mathrm{s})\right) $$ $$ C_{\mathrm{P}}(0.1 \mu \mathrm{m}, 30 \mathrm{s})\approx 8.08 \times 10^{18} \mathrm{~kmol/m}^{3} $$ Now we need to find the mole fraction of \(P\).

05

Compute mole fraction of \(P\)

To find the mole fraction of \(P\), we can use the following formula: $$ x_{\mathrm{P}} = \frac{C_{\mathrm{P}}(x, t)}{C_{\mathrm{P}}(x, t) + C_{\mathrm{Si}}(x, t)} $$ We are given the mass densities and molecular weights for both \(P\) and \(Si\). We can use these values to find their molar concentrations \(C_{\mathrm{P}}(x, t)\) and \(C_{\mathrm{Si}}(x, t)\): - \(C_{\mathrm{Si}} = \frac{\rho_{\mathrm{Si}}}{M_{\mathrm{Si}}} \approx \frac{2300 \mathrm{~kg} / \mathrm{m}^{3}}{28.09 \mathrm{~kg} / \mathrm{kmol}} \approx 8.19 \times 10^{4} \mathrm{~kmol/m}^{3}\) Now, we can calculate the mole fraction of \(P\): $$ x_{\mathrm{P}} = \frac{8.08 \times 10^{18}}{8.08 \times 10^{18} + 8.19 \times 10^{4}} \approx 0.999 $$ So the mole fraction of phosphorus at a depth of \(0.1 \mu \mathrm{m}\) in the silicon after \(30 \mathrm{~s}\) is approximately \(0.999\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diffusion

Diffusion is the process through which particles spread from regions of high concentration to regions of lower concentration. This process is driven by the random motion of particles, which continues until there is a uniform distribution. In our exercise, the diffusion of phosphorus (P) into silicon (Si) is under investigation. When a Si wafer is introduced into a high-temperature environment, the diffusion process accelerates due to increased kinetic energy of the molecules.
This exercise uses an equation to describe how the concentration of P changes over time and depth in the Si. The equation has an exponential component, indicating that concentration decreases as the square of the distance from the surface increases. This is analogous to the way heat spreads through a medium, governed by a similar mathematical expression.

Temperature Distribution

In the context of heat transfer, temperature distribution describes how heat spreads across a material. It provides insight into how temperature varies with both time and position within a material. In the given problem, the temperature distribution follows a similar formula as the concentration distribution for P in Si. The formula includes the term \(T(x, t) - T_i\), showing the temperature difference from the initial temperature \(T_i\).
As time proceeds, the energy spreads out, described by an exponential term.
This indicates that closer areas reach the new temperature state faster than those deeper within the material due to the nature of the exponential decay term related to space.

Thermal Diffusivity

Thermal diffusivity ( \(\alpha\)) is a material-specific parameter that measures the rate at which heat spreads through a material. It is defined as the ratio of thermal conductivity to the product of density and specific heat capacity. Essentially, higher thermal diffusivity means a material can transfer heat more quickly.
In the given problem, the analogous parameter for mass transfer is the diffusion coefficient (\(D_{P-Si}\)) for P into Si. Similar to thermal diffusivity, the diffusion coefficient dictates how rapidly P atoms disperse within the Si wafer.
A higher value would mean P diffuses more rapidly through the Si, akin to how higher thermal diffusivity allows for faster heat transfer. It's crucial in designing processes where controlled distribution is needed, whether for heat or mass.

Concentration Distribution

Concentration distribution describes how the concentration of a substance varies within a medium. This is a critical aspect of mass transfer, similar to how temperature distribution works in heat transfer.In our exercise, the concentration distribution of phosphorus on a Si wafer is modeled by an expression with parameters analogous to those used for temperature distribution.
\( C_P(x,t)\) represents how the concentration of P changes with depth (\(x\)) and time (\(t\)). The formula illustrates how the initial concentration spreads over time due to diffusion.
Such distribution models allow engineers and scientists to predict how materials will behave under certain conditions, such as high temperatures, making them pivotal in the design and optimization of semiconductor manufacturing processes.