Question

For the Maxwell-Boltzmann speed distribution, the ratio of the root-mean-square speed (v_rms) and the most probable speed (v_max) is
Given : Maxwell-Boltzmann speed distribution function for a collection of particles of mass m is
\(f(v)=(\frac{m}{2\pi k_BT})^{3/2}4\pi v^2\text{exp}(-\frac{mv^2}{2k_BT})\)
where, v is the speed and k_BT is the thermal energy.

• A. \(\sqrt{\frac{3}{2}}\)
• B. \(\sqrt{\frac{2}{3}}\)
• C. \(\frac{3}{2}\)
• D. \(\frac{2}{3}\)

Answer 1

The Correct Option is A

To determine the ratio of the root-mean-square speed (\(v_{\text{rms}}\)) and the most probable speed (\(v_{\text{max}}\)) for particles following the Maxwell-Boltzmann speed distribution, we need to derive both speeds from the given distribution and then compute their ratio.

The Maxwell-Boltzmann speed distribution function is given by:

\(f(v) = \left(\frac{m}{2\pi k_B T}\right)^{3/2} 4\pi v^2 \exp\left(-\frac{mv^2}{2k_B T}\right)\)

where:

• \(v\) is the speed of particles,
• \(m\) is the particle mass,
• \(k_B\) is the Boltzmann constant,
• \(T\) is the temperature.

1. **Root-Mean-Square Speed \((v_{\text{rms}})\):**

The root-mean-square speed is given by the formula:

\(v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}\)

2. **Most Probable Speed \((v_{\text{max}})\):**

The most probable speed is derived by maximizing the function \(f(v)\). Setting the derivative \( \frac{d}{dv}f(v) = 0\), yields:

\(v_{\text{max}} = \sqrt{\frac{2k_B T}{m}}\)

3. **Ratio of \(v_{\text{rms}}\) to \(v_{\text{max}}\):**

To find the ratio:

\[ \frac{v_{\text{rms}}}{v_{\text{max}}} = \frac{\sqrt{\frac{3k_B T}{m}}}{\sqrt{\frac{2k_B T}{m}}} = \sqrt{\frac{3}{2}} \]

Thus, the correct answer is \(\sqrt{\frac{3}{2}}\).

This answer matches option one: \(\sqrt{\frac{3}{2}}\), and hence it is the correct choice.