Question

Calculate the rms speed in ${\rm{cm/sec}}$ at $25^\circ {\rm{C}}$ of a free electron.

Answer 1

Hint:
The formula for rms speed as per Maxwell-Boltzmann distribution of molecular speed can be used along with the given condition of temperature.

Complete step by step solution
We know that the molecular speed is not same for all the molecules in a gas and this led to Maxwell-Boltzmann distribution of molecular speeds as per which different speeds are given by the following formulae:

- Most probable speed: most of the gaseous molecules have this speed and the formula is:
${v_{mp}} = \sqrt {\dfrac{{2{k_B}T}}{m}} $
Here, ${k_B}$ is the Boltzmann constant, $T$ is the temperature of the gas and $m$ is the molecular mass.
- Average speed: it is the average speed for all the molecules and the formula is:
${v_{av}} = \sqrt {\dfrac{{8{k_B}T}}{\pi }} $
- Root mean square speed: It is also known as rms speed and as the name indicates it is determined by taking square root of the mean of the squared speeds and its formula is:
${v_{rms}} = \sqrt {\dfrac{{3{k_B}T}}{m}} $

Here, we have to determine the rms speed for a free electron at $25^\circ {\rm{C}}$. Let’s first convert the given temperature to absolute scale as follows:
$
T = \left( {25 + 273} \right)\;{\rm{K}}\\
 = {\rm{298}}\;{\rm{K}}
$

Now, we can calculate the rms speed by substituting ${m_e} = 9.11 \times {10^{ - 31}}{\rm{\;kg}}$, ${k_B} = 1.38 \times {10^{ - 23}}\;{{\rm{m}}^{\rm{2}}} \cdot {\rm{kg}} \cdot {{\rm{s}}^{ - 2}} \cdot {{\rm{K}}^{ - 1}}$ and $T = {\rm{298}}\;{\rm{K}}$ in the above written formula for rms speed as follows:
$
{v_{rms}} = \sqrt {\dfrac{{3\left( {1.38 \times {{10}^{ - 23}}\;{{\rm{m}}^{\rm{2}}} \cdot {\rm{kg}} \cdot {{\rm{s}}^{ - 2}} \cdot {{\rm{K}}^{ - 1}}} \right)\left( {{\rm{298}}\;{\rm{K}}} \right)}}{{9.11 \times {{10}^{ - 31}}{\rm{\;kg}}}}} \\
 = {\rm{11}}{\rm{.64}} \times {10^4}\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}
$

As we are required to report the value in ${\rm{cm/sec}}$, so let’s write the conversion factor for distance as follows:
$\dfrac{{100\;{\rm{cm}}}}{{1\;{\rm{m}}}}$
Finally, we can now convert the units of the calculated rms speed by using the above conversion factor as follows:
$
{v_{rms}} = \left( {\dfrac{{100\;{\rm{cm}}}}{{1\;{\rm{m}}}}} \right) \times {\rm{11}}{\rm{.64}} \times {10^4}\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}\\
 = {\rm{11}}{\rm{.64}} \times {10^6}\;{\rm{cm}} \cdot {{\rm{s}}^{ - 1}}
$

Hence, the rms speed of a free electron at $25^\circ {\rm{C}}$ is ${\rm{11}}{\rm{.64}} \times {10^6}\;{\rm{cm}} \cdot {{\rm{s}}^{ - 1}}$.

Note:
At times we also take the formula of molecular speeds in terms of universal gas constant and molar mass as Boltzmann constant and universal gas constant are related by Avogadro constant only.