A lottery machine uses blowing air to keep 2000 Ping-Pong balls bouncing around inside a $1.0\mathrm{m}\times 1.0\mathrm{m}\times 1.0\mathrm{m}$box. The diameter of a Ping-Pong ball is $3.0\mathrm{cm}$. What is the mean free path between collisions? Give your answer in$\mathrm{cm}$.

Number of Ping-Pong balls in a lottery machine$=2000$

Lottery machine dimensionlocalid="1648284969427" $=1.0\mathrm{m}\times 1.0\mathrm{m}\times 1.0\mathrm{m}$

The diameter of a Ping-Pong ball$=3.0cm$

Due to collisions with other molecules, a molecule undergoing distance travel experiences a delay between diffusing to a different position due to these collisions.

The molecules need a certain amount of time to diffuse to a new position.

In equation (20.3), the mean free path $\lambda$is the distance the molecules travel between collisions on average.

$\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}\to \left(1\right)$

$r=$radius of the particle .

The number density is given by equation,

role="math" localid="1648285510457" $\text{number density}=\frac{N}{V}\to \left(2\right)$

role="math" localid="1648285435990" $N=2000$number of particles

$V=$volume of the box.

To calculate the volume,

$V=1.0\mathrm{m}\times 1.0\mathrm{m}\times 1.0\mathrm{m}=1.0{\mathrm{m}}^3$

Use equation (2) to get number density,

$\frac{N}{V}=\frac{2000}{1.0{\mathrm{m}}^3}=2000{\mathrm{m}}^{-3}$

Diameter of the ball $d=3\mathrm{cm}$,

Then the radius will be,$r=\frac{d}{2}=\frac{3\mathrm{cm}}{2}=1.5\mathrm{cm}$

Put the values for $(N/V)$and $r$into equation (1) to get $\lambda$,

$\begin{array}{r}\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}\end{array}$

$\begin{array}{r}=\frac{1}{4\sqrt{2}\pi \left(2000{\mathrm{m}}^{-3}\right)(0.015\mathrm{m}{)}^2}\end{array}$

role="math" localid="1648285716549" $=12.5\times {10}^{-2}\mathrm{m}$

$=12.5\mathrm{cm}$

Hence, the mean free path between collision is$\lambda =12.5\mathrm{cm}.$