一家晶圆制造公司使用悬浮区熔法来提纯多晶硅锭，每个硅锭具有均匀的横截面积，长度为 200 cm。该方法包括将杂质浓度为 $C_0$ atoms/$\text{cm}^3$ 的多晶硅锭放置在单个籽晶的顶部。然后通过射频（RF）线圈对多晶硅锭进行外部加热，从而使硅锭局部熔化。杂质更倾向于留在熔融态而非固态中，因此随着线圈和熔区向上移动，一个更纯的单晶在籽晶顶部凝固。该方法的原理示意图如图 1 和图 2 所示。 对于长度为 $L$ cm 的“悬浮”熔区，熔区中的杂质浓度 $C$ atoms/$\text{cm}^3$ 与射频线圈移动的距离 $x$ cm 满足以下微分方程：
$\frac{\text{d}C}{\text{d}x} = \frac{1}{L}(C_0 - kC),$
其中 $k$ 是一个常数，满足 $0 < k < 1$。 该公司采用的“悬浮”熔区长度 $L$ cm 为 2 cm，且 $0 \le x \le 198$。已知当 $x = 0$ 时，$C = C_0$。

1. Substitute the given value of $L$ into the differential equation and separate the variables.

Given $L = 2$, the differential equation becomes:
$\frac{\text{d}C}{\text{d}x} = \frac{1}{2}(C_0 - kC)$
Separating the variables $C$ and $x$:
$\int \frac{1}{C_0 - kC} \, \text{d}C = \int \frac{1}{2} \, \text{d}x$

2. Integrate both sides to find the general solution.

$-\frac{1}{k} \ln |C_0 - kC| = \frac{x}{2} + A$

3. Apply the initial condition to find the constant of integration, $A$.

Given that when $x = 0$, $C = C_0$:
$-\frac{1}{k} \ln |C_0 - kC_0| = \frac{0}{2} + A$
Since $0 < k < 1$ and $C_0 > 0$, we have $C_0 - kC_0 = C_0(1 - k) > 0$. Therefore:
$A = -\frac{1}{k} \ln [C_0(1 - k)]$
Substituting $A$ back into the equation:
$-\frac{1}{k} \ln |C_0 - kC| = \frac{x}{2} - \frac{1}{k} \ln [C_0(1 - k)]$

4. Rearrange the equation to make $C$ the subject.

Multiply the entire equation by $-k$:
$\ln |C_0 - kC| = -\frac{kx}{2} + \ln [C_0(1 - k)]$
$\ln |C_0 - kC| - \ln [C_0(1 - k)] = -\frac{kx}{2}$
$\ln \left| \frac{C_0 - kC}{C_0(1 - k)} \right| = -\frac{kx}{2}$
Since $C_0 - kC$ is positive initially and continuous, we can drop the absolute value:
$\frac{C_0 - kC}{C_0(1 - k)} = e^{-\frac{kx}{2}}$
$C_0 - kC = C_0(1 - k)e^{-\frac{kx}{2}}$
$kC = C_0 - C_0(1 - k)e^{-\frac{kx}{2}}$
$C = \frac{C_0}{k} \left[ 1 - (1 - k)e^{-\frac{kx}{2}} \right]$

1. Identify the initial point and horizontal asymptote from the solution for $C$.

From part (a), the solution is $C = \frac{C_0}{k} \left( 1 - (1-k) e^{-\frac{kx}{2}} \right)$.
When $x = 0$, $C = C_0$.
As $x \to \infty$, $e^{-\frac{kx}{2}} \to 0$ (since $k > 0$).
Thus, $C \to \frac{C_0}{k}$. The horizontal asymptote is $C = \frac{C_0}{k}$.

2. Determine the shape of the curve based on the first derivative.

From the given differential equation, $\frac{\text{d}C}{\text{d}x} = \frac{1}{2}(C_0 - kC)$.
Since $0 < k < 1$ and $C$ starts at $C_0$, the initial gradient is $\frac{1}{2}C_0(1-k) > 0$.
As $C$ increases towards $\frac{C_0}{k}$, the term $(C_0 - kC)$ remains positive but decreases towards $0$.
Therefore, the gradient is positive but strictly decreasing, meaning the curve is increasing and concave downwards.

3. Sketch the graph incorporating the valid domain.

Draw the horizontal $x$-axis and vertical $C$-axis.
Plot the $C$-intercept at $(0, C_0)$.
Draw a horizontal dotted line at $C = \frac{C_0}{k}$ (note that $\frac{C_0}{k} > C_0$ since $0 < k < 1$).
Draw a smooth, increasing, concave down curve from the intercept, approaching the asymptote.
Terminate the graph at $x = 198$, matching the domain given in the context.

1. Briefly solve the differential equation from part (a) to express $C$ in terms of $x$, as the result is required.

As required from part (a), we first find the expression for $C$ by separating variables and integrating:
$\int \frac{1}{C_0 - kC} \,\text{d}C = \int \frac{1}{2} \,\text{d}x$
$-\frac{1}{k} \ln |C_0 - kC| = \frac{x}{2} + A$
Using the initial condition $C = C_0$ when $x = 0$:
$A = -\frac{1}{k} \ln(C_0 - kC_0) = -\frac{1}{k} \ln(C_0(1-k))$
$\ln\left(\frac{C_0 - kC}{C_0(1-k)}\right) = -\frac{kx}{2} \implies C = \frac{C_0}{k}\left(1 - (1-k)e^{-\frac{kx}{2}}\right)$

2. Substitute $C = 2C_0$ and $k = 0.3$ to find the distance $x$ moved.

$2C_0 = \frac{C_0}{0.3}\left(1 - (1 - 0.3)e^{-\frac{0.3x}{2}}\right)$
$0.6 = 1 - 0.7e^{-0.15x}$
$0.7e^{-0.15x} = 0.4 \implies e^{-0.15x} = \frac{4}{7}$
$x = -\frac{1}{0.15}\ln\left(\frac{4}{7}\right) = \frac{20}{3}\ln 1.75 \approx 3.7308 \text{ cm}$

3. Convert the speed to consistent units and calculate the time taken $t$.

The distance $x$ is in cm, so the upward speed must be in cm/h:
$v = 8 \text{ mm/h} = 0.8 \text{ cm/h}$
$t = \frac{x}{v} = \frac{3.7308}{0.8} \approx 4.6635 \text{ hours}$
So the time taken is approximately $4.66$ hours.

4. Set up the relationship for the rate of change of $C$ with respect to time using the chain rule.

$\frac{\text{d}C}{\text{d}t} = \frac{\text{d}C}{\text{d}x} \times \frac{\text{d}x}{\text{d}t}$
We know $\frac{\text{d}x}{\text{d}t} = 0.8 \text{ cm/h}$ and the differential equation gives $\frac{\text{d}C}{\text{d}x} = \frac{1}{2}(C_0 - kC)$.

5. Evaluate the rate of change of concentration at the instant when $C = 2C_0$.

At the instant when $C = 2C_0$ and $k = 0.3$:
$\frac{\text{d}C}{\text{d}x} = \frac{1}{2}(C_0 - 0.3(2C_0)) = \frac{1}{2}(C_0 - 0.6C_0) = 0.2C_0$
$\frac{\text{d}C}{\text{d}t} = (0.2C_0) \times 0.8 = 0.16C_0$

1. Concisely solve the differential equation from part (a) to find the expression for $C$.

Separating variables for the given differential equation:
$\int \frac{1}{C_0 - kC} \, \mathrm{d}C = \int \frac{1}{L} \, \mathrm{d}x$
$-\frac{1}{k} \ln(C_0 - kC) = \frac{x}{L} + A$
Using the initial condition $C = C_0$ when $x = 0$:
$A = -\frac{1}{k} \ln(C_0(1-k))$
$\ln \left( \frac{C_0 - kC}{C_0(1-k)} \right) = -\frac{kx}{L} \implies C_0 - kC = C_0(1-k)\mathrm{e}^{-\frac{kx}{L}}$
$C = \frac{C_0}{k} \left[ 1 - (1-k)\mathrm{e}^{-\frac{kx}{L}} \right]$

2. Analyze the effect of a shorter melt zone length $L$ on the impurity concentration $C$.

From the expression for $C$, for a shorter length $L$, the value of $\frac{kx}{L}$ is larger, making $\mathrm{e}^{-\frac{kx}{L}}$ smaller.
Since $1-k > 0$, the term $1 - (1-k)\mathrm{e}^{-\frac{kx}{L}}$ becomes larger.
Therefore, a shorter length $L$ results in a larger $C$ for any given $x > 0$.
(Alternatively, directly from $\frac{\mathrm{d}C}{\mathrm{d}x} = \frac{1}{L}(C_0 - kC)$, a smaller $L$ yields a larger positive $\frac{\mathrm{d}C}{\mathrm{d}x}$, meaning $C$ increases more rapidly.)

3. Relate the higher concentration $C$ to the purity of the solidifying crystal to draw a conclusion.

A larger $C$ means a higher concentration of impurities in the melt zone.
Because the newly formed single crystal draws its material from this melt zone, a higher impurity concentration in the melt leads to a higher concentration of impurities solidifying into the crystal.
As this makes the single crystal less pure, a shorter length is not preferable.