Question

A pure Silicon wafer is doped with Boron by exposing it to $B_2O_3$ vapour at an elevated temperature. It takes 1000 seconds to reach a Boron concentration of $10^{20}$ atoms $m^{-3}$ at a depth of $1 \, \mu m$. The time taken to reach the same concentration of Boron at a depth of $2 \, \mu m$ is (in seconds): ................ (rounded off to nearest integer).

Hint:

In constant-source diffusion, the relationship is $x \propto \sqrt{t}$. Hence, if depth is doubled, the time must increase by a factor of 4. Always check whether it’s a {constant-source} or {limited-source} case before applying this proportionality.

Answer 1

Correct Answer: 3840

Step 1: Recall the governing principle.
Diffusion in solids under constant surface concentration conditions is described by Fick’s second law. The general solution gives: \[ C(x,t) = C_s \, \text{erfc} \left( \frac{x}{2\sqrt{Dt}} \right) \] Here, $C_s$ is the surface concentration, $D$ is diffusivity, $x$ is depth, and $t$ is time. For a fixed concentration at depth $x$, the term $\frac{x}{\sqrt{t}}$ must remain constant.
Step 2: Establish proportionality relation.
Thus: \[ \frac{x_1}{\sqrt{t_1}} = \frac{x_2}{\sqrt{t_2}} \Rightarrow t_2 = t_1 \left( \frac{x_2}{x_1} \right)^2 \]
Step 3: Substitute known values.
\[ t_1 = 1000 \, s, x_1 = 1 \, \mu m, x_2 = 2 \, \mu m \] \[ t_2 = 1000 \times \left( \frac{2}{1} \right)^2 \] \[ t_2 = 1000 \times 4 = 4000 \, s \]
Step 4: Interpretation.
This means that to reach the same Boron concentration at twice the depth, the diffusion process requires four times the time, since depth grows with $\sqrt{t}$. Final Answer: \[ \boxed{4000 \, \text{seconds}} \]