The probability of finding a particle in the speed interval v to v + dv is given by:

$$P (v) = 4 \pi v^{2} \left( \frac{m}{2 \pi k_{B} T} \right)^{\frac{3}{2}} e^{- \frac{m v^{2}}{2 k_{B} T}}$$

, where

- $4 \pi v^{2}$ is the density of states
- $- \frac{m v^{2}}{2 k_{B} T}$ is the ratio of kinetic energy to thermal energy
- $\left( \frac{m}{2 \pi k_{B} T} \right)^{\frac{3}{2}} e^{- \frac{m v^{2}}{2 k_{B} T}}$ is the Boltzmann distribution

a) P(v) = 0 at v = 0 and as v tends to infinity.

b) P(v) reaches a maximum at some intermediate value of v.

– The position of this peak and the shape of the distribution function varies with temperature: As temperature increases, the peak occurs at higher v and the width of the distribution becomes broader: the average speed increases and a larger fraction of molecules can attain very high speeds.

**Various measures of the characteristic speed**

– v_{mp} is the most probable speed at the peak of the distribution function.

– v_{av} is the average speed of the distribution.

– Note: v_{rms} > v_{av} > v_{mp}

**The most probable speed **

The most probable speed occurs at:

$$\frac{d P(v)}{dv} = 0$$

hence,

$$v_{mp} = \sqrt{\frac{2 k_{B} T}{m}}$$

**The average speed**

The average speed is given by:

$$\bar{v} = \int\limits_{0}^{\infty} P(v) \, v \, dv$$

hence,

$$\bar{v} = \sqrt{\frac{8 k_{B} T}{\pi m}}$$

**The root-mean-square speed**

The root-mean-square speed is given by:

$$v_{\text{rms}} = \left( \int\limits_{0}^{\infty} P(v) \, v^{2} \, dv \right)^{\frac{1}{2}}$$

hence,

$$v_{\text{rms}} = \sqrt{\frac{3 k_{B} T}{m}}$$

Next: Theorem Of Equipartition Of Energy