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Scenario

Imagine you randomly spread in a cubic space of volume $m^3$ (or in a sphere of volume $V_s$, as you prefer), $n_A$ particles $A$ and $n_B$ particles $B$. Particles $A$ are spheres of volume $V_A$ and are not moving. Particles $B$ have no volume (they are points) and are moving linearly (reflecting at the boundaries of the space (cube or sphere)) at a constant speed $s$.

We can assume that when a particle $B$ meet a particle $A$ (collision), we count a collision but the particle $B$ just go through the particle $A$. Whatever is the model that is easier to calculate.

Question

What function describes the probability that there is $x$ collisions in a time interval $\Delta t$. Or what is the expected the number of collisions?

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  • $\begingroup$ If A particles are at fixed points in space, the answer will depend critically on their spatial distribution. $\endgroup$ – user41033 Aug 2 '17 at 8:24
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If you want a number of collisions in a given time $\Delta t$, then simply from a units perspective, the other answer doesn't make sense, as it produces a number with dimensions of length. You want something that is dimensionless.

Often, physical chemistry textbooks will provide a derivation for collisions per unit volume per unit time. The proofs are a little bit involved and I'd rather refer you to a textbook instead of typing it all out. However, the bottom line is that the collision frequency is given by

$$Z_{AB} = \pi(r_A + r_B)^2\left[\langle v_A \rangle^2 + \langle v_B \rangle^2 \right]^{1/2} \left(\frac{N_A}{V}\frac{N_B}{V}\right)$$

  • $Z_{AB}$ is the number of collisions between A and B per unit time per unit volume
  • $r_A$ and $r_B$ are the radii of A and B
  • $\langle v_A \rangle^2$ and $\langle v_B \rangle^2$ are the mean square velocities of A and B
  • $N_A$ and $N_B$ are the number of particles of A and B
  • $V$ is the total volume

For a derivation of this I suggest reading Levine's Physical Chemistry, 6th ed, section 14.7.

For your specific case you've assumed that $r_B = 0$, $v_A = 0$, $v_B = s$. Therefore this simplifies to

$$Z_{AB} = \pi r_A^2 \cdot s \cdot \left(\frac{N_A}{V}\frac{N_B}{V}\right)$$

Obviously, the radius of A, $r_A$ is related to its volume $V_A$ by $V_A = 4\pi r_A^3 /3$. To obtain the number of collisions in a given time you just need to multiply by the total volume and total time.

$$f = \pi r_A^2 \cdot s\Delta t \cdot \left(\frac{N_A N_B}{V}\right)$$

As you can see, the form of this equation is not all too different from the one in the other answer. The key difference is that the distance travelled $s\Delta t$ must be multiplied by a cross-sectional collision area, equal to $\pi r_A^2$, and not by a volume. The textbooks will explain this well, I will try to elaborate on this later if I have time.

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With assumption that $n_A$ and $n_B$, i.e., the number of particles $A$ and $B$, are large and are in random motion, then the probability of a particle $B$ to be found within the volume of particles $A$ is $$\frac{n_A V_A}{V_s}$$

This will hold for every particle $B$. So, when each particle moves $s \Delta t$ distance, the number of collisions it will undergo is

$$\frac{s\Delta t\cdot n_A V_A}{V_s}$$

For $n_B$ particles the number of collisions is

$$\frac{n_B \cdot s \Delta t \cdot n_A V_A}{V_s}$$

With my assumption at top, it will either approach 1 or 0.

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  • $\begingroup$ I believe this is not entirely correct. The distance $s\Delta t$ must be multiplied by a cross-sectional area, not a volume. Also I don't see why the number of collisions must be 1 or 0. $\endgroup$ – orthocresol Aug 2 '17 at 4:36

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