Question

Component A diffuses into a solid to a depth of 10 μm in 1 hour at 300 K. Treat diffusion in one dimension. Find the time (in seconds) for A to diffuse to the same depth at 600 K. (Round off to 1 decimal). Diffusivity is given by: \[ D_A = D_A^0 \exp\left( -\frac{E_a}{k_B T} \right) \] Given: \[ E_a = 0.3 \text{ eV}, \quad k_B = 8.62\times10^{-5}\,\text{eV/K} \] Depth is same → use relation: \[ x \sim \sqrt{Dt} \Rightarrow Dt = \text{constant} \] Thus: \[ \frac{t_2}{t_1} = \frac{D_1}{D_2} \]

Hint:

For equal diffusion depth, use \( x \sim \sqrt{Dt} \Rightarrow Dt = \text{constant} \). Higher temperature increases diffusivity exponentially.

Answer 1

Correct Answer: 9.7

Diffusivities at 300 K and 600 K:
\[ D_1 = D_0 \exp\left( -\frac{0.3}{8.62\times10^{-5}\times 300} \right) \] \[ D_2 = D_0 \exp\left( -\frac{0.3}{8.62\times10^{-5}\times 600} \right) \] Exponent values:
\[ \frac{0.3}{8.62\times10^{-5}\times 300} \approx 11.60 \] \[ \frac{0.3}{8.62\times10^{-5}\times 600} \approx 5.80 \] Thus: \[ \frac{D_1}{D_2} = \exp(-(11.60 - 5.80)) = \exp(-5.80) \] \[ \frac{D_1}{D_2} \approx 0.0030 \] Since: \[ t_1 = 1\text{ hr} = 3600\text{ s} \] \[ t_2 = t_1 \frac{D_1}{D_2} \] \[ t_2 = 3600 \times 0.0030 = 10.8\text{ s} \] Rounded to one decimal: \[ t_2 = 10.8\text{ s} \]