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.


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?

  • $\begingroup$ If A particles are at fixed points in space, the answer will depend critically on their spatial distribution. $\endgroup$
    – user41033
    Commented Aug 2, 2017 at 8:24

2 Answers 2


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.


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.

  • $\begingroup$ 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. $\endgroup$ Commented Aug 2, 2017 at 4:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.