Time for a particle undergoing brownian motion to reach a point in a volume I was wondering how one could calculate the average time a particle needs to reach a random point in a small sphere (filled by water) with a radius of maybe $10 \mu m$. I thought of using the Stokes-Einstein-Equation, but then I just get a diffusion coefficient with the unit [m²/s]. Does anyone have an idea on how to solve this?
 A: First remark, the average hitting time is finite because the volume is finite. None of what I write would make sense in an infinite system. 
Let us consider that the target is a ball of radius $a$ at the center of the sphere and let us call $T(\vec r)$ the average hitting time for a Brownian particle starting at position $\vec r$ from the origin. $T(\vec r)$ depends only on $r=\|\vec r\|$. Consider now a Brownian motion $\vec B$ during a short time $\mathrm dt$ and compute the variation of the hitting time 
$$\mathrm dt=T(\vec r)-T(\vec r+\vec B_{\mathrm dt}).$$
Use the Taylor expansion in spherical coordinates of the right-hand side
$$\mathrm dt=-\vec\nabla T(\vec r)\cdot\vec B_{\mathrm dt}-\frac12\vec B_{\mathrm dt}\cdot H_T(\vec r)\cdot \vec B_{\mathrm dt}$$
where $H_T$ is the Hessian matrix of $T$ which contains only one non-zero element equal to $\frac1{r^2}\partial_r(r^2\partial_r T(r))$ on the diagonal. Taking the average over all realisations of the Brownian motion, one gets
$$\mathrm dt=-\frac12\frac1{r^2}\partial_r\left(r^2\partial_r T(r)\right) \; 2D\mathrm dt,$$ or in a simpler form, with $\Delta_r$ denoting the spherical Laplacian,
$$D\Delta_r T(r)=-1.\tag{1}$$
This form is quite general for the average hitting time. It is actually the Fokker-Planck equation in which we have replaced the time derivative by $-1$. 
Let us solve (1). We have $$T(r)=-\frac{r^2}{6D}+\frac{A}r+B.$$
$A$ and $B$ are constants. We must have $T(a)=0$. The second condition depends on the boundary at $r=R$. Let us consider that the boudary is reflecting, so the particles bounce on the sphere and continue to diffuse inside. This is a von Neumann boundary condition, that translates mathematically into $\left.\partial_rT(r)\right\rvert_{r=R}=0$. This defines $A=-R^3/3D$. With $T(a)=0$ we find
$B=a^2/6D+R^3/3Da$. 
Finally 
$$T(r)=\frac{a^2-r^2}{6D}+\frac{R^3}{3D}\left(\frac1a-\frac1r\right).$$
If the starting point is uniformally distributed inside the sphere of radius $R$, the average is 
$$ \int_a^R\frac{3r^2}{R^3-a^3}T(r)\mathrm dr=\frac{(R-a)^2}{15Da}\frac{5R^3+6 R^2a + 3 Ra^2 + a^3}{R^2+Ra+a^2}.$$
Therefore, for $a\ll R$ we get $$\left\langle T\right\rangle\approx\frac{R^3}{3Da}.$$ 
The average hitting time scales like $R^3$, so is actually proportional to the volume and inversely proportional to the radius of the target. 
A: Note that a particle can pass through a particular location many many time, what you are looking for is the first passage time, FPT $F(r',t|r)$, which is the probability of taking time $t$ to move from point $r$ to point $r'$. The quantity you are looking for is the mean FPT, given
$$ \langle T \rangle = \int_0^\infty t F(r',t|r) dt $$
In general free space (particular $d\ge3$), this mean time may can be infinite because the particle may never pass a point. For confined dominate, it will eventually does, which means that $\langle T \rangle$ is finite.
In confined domain, the qualitative result should depend on the distance of $r,r'$ from the spherical wall as well as distance $|r-r'|$. I don't think there is any simple solution.
For your reaction type system, it may be easier to just do the particle simulation and calculate $\langle T \rangle$. Alternatively, the calculation of $\langle T \rangle$ in random walk problem is equivalent to solving diffusion equation with an absorbing boundary condition, or a sink at $r'$. The total probability will decrease over time and the FPT is simply the $F(r',t) = \partial_t \int dr P(r,t)$. This can be done on existing software to solve diffusion equation, so you can see the result quickly.

Mathematically, the FPT is related with the solution (or Green function) $P(r',t|r)$ of the diffusion equation (subjected to the boundary condition with diffusion constant $D$) as follow:
$$ P(r',t|r) = \int_0^\infty F(r',t'|r) P(r',t-t'|r') dt'$$
The relationship is clear as the walker need to go to spend $t'$ to go to the point $r'$ first and then take another $t-t'$ to loop back. Once $P$ is known, the $F$ can be calculated by Laplace transform and so does $\langle T \rangle$.
