Given:
Time = 8.30 s ahead of the hemoglobin
For the equation of both glucose and hemoglobin:
[tex]\begin{gathered} \sqrt{2D_gt_g}=\sqrt{2D_h(t_g+\Delta t)} \\ \\ D_gt_g=D_h(t_g+\Delta t) \\ \\ D_gt_g=D_ht_g+D_h\Delta_t \end{gathered}[/tex]Rewrite the equation for tg:
[tex]\begin{gathered} D_gt_g-D_ht_g=D_h\Delta_t \\ \\ t_g(D_g-D_h)=D_h\Delta_t \\ \\ t_g=\frac{D_h\Delta_t}{D_g-D_h} \\ \\ \end{gathered}[/tex]Where:
Dh is the diffusion coefficient of hemoglobin = 6.9 x 10ā»Ā¹Ā¹ mĀ²/s
Dg is the diffusion coefficient of glucose = 6.7 x 10ā»Ā¹ā° mĀ²/s
Īt = 8.30 seconds
Substitute these values for the variables in the equation and solve:
[tex]\begin{gathered} t_g=\frac{6.9\times10^{-11}*8.30}{6.7\times10^{-10}-6.9\times10^{-11}} \\ \\ t_g=\frac{5.727\times10^{-10}}{6.01\times10^{-10}} \\ \\ t_g=0.953\text{ seconds} \end{gathered}[/tex]Therefore, the time that passes before the glucose is 8.30 s ahead of the hemoglobin is 0.953 seconds.
ANSWER:
0.953 seconds
