Question

The value closest to the thermal velocity of a Helium atom at room temperature ${\text{(300}}\,{\text{K)}}$ in ${\text{m}}{{\text{s}}^{ - {\text{1}}}}$ is:
$\left[ {{k_{\text{B}}} = 1.4 \times {{10}^{ - 23}}\,{\text{J}}{{\text{K}}^{ - 1}}:\,\,{m_{{\text{He}}}} = 7 \times {{10}^{ - 27}}\,{\text{kg}}} \right]$
(A) $1.3 \times {10^4}$
(B) $1.3 \times {10^3}$
(C) $1.3 \times {10^5}$
(D) $1.3 \times {10^2}$

Answer 1

In order to solve this problem we are going to use the concept of root mean square velocity of the gas molecules which is nothing but the root mean square of the individual gas molecule’s velocity.After applying its formula we can arrive at the desired result.

Formula used:
${V_{{\text{rms}}}} = \sqrt {\dfrac{{3RT}}{M}}$ ……….(1)
Where,
${V_{{\text{rms}}}}$ indicates the RMS velocity.
$R$ indicates gas constant.
$T$ indicates temperature in kelvin scale, and
$M$ indicates the molecular mass of the gas.
${k_B} = \dfrac{R}{{{N_{\text{A}}}}}$ ………… (2)
Where,
${k_B}$ indicates Boltzmann constant.
$R$ indicates gas constant, and
${N_{\text{A}}}$ indicates Avogadro’s number.
${N_{\text{A}}} = \dfrac{{{M_{{\text{He}}}}}}{{{m_{{\text{He}}}}}}$ ………..(3)
${M_{{\text{He}}}}$ indicates molecular mass of the gas.
${m_{{\text{He}}}}$ indicates mass of one atom.

Complete step by step answer:
Thermal velocity or thermal velocity is a normal velocity in the thermal motion of particles containing a gas, liquid, etc. Thus, literally, a temperature measurement is thermal velocity.
We know that root mean square (RMS) velocity of molecules is given by the equation(1).But in the problem, we are supplied with Boltzmann constant and mass of one atom of Helium, so need to modify the equation (1)
Now, we use equation (3) in equation (2):

$${k_B} = \dfrac{R}{{{N_{\text{A}}}}} \\\ \Rightarrow{k_B} = \dfrac{R}{{\left( {\dfrac{{{M_{{\text{He}}}}}}{{{ m _{{\text{He}}}}}}} \right)}} \\\ \Rightarrow{k_B} = \dfrac{{R{ m _{{\text{He}}}}}}{{{M_{{\text{He}}}}}} \\\ \Rightarrow\dfrac{{{k_B}}}{{{m_{{\text{He}}}}}} = \dfrac{R}{{{M_{{\text{He}}}}}} \\\$$

Now, we substitute:
$\dfrac{{{k_B}}}{{{m_{{\text{He}}}}}} = \dfrac{R}{{{M_{{\text{He}}}}}}$ in equation (1):

\Rightarrow{V_{{\text{rms}}}} = \sqrt {\dfrac{{3RT}}{{{M_{{\text{He}}}}}}} \\\ \Rightarrow{V_{{\text{rms}}}}= \sqrt {\dfrac{{3{k_{\text{B}}}T}}{{{m_{{\text{He}}}}}}} \\\ $$ …… (4) Finally, substitute, $${k_{\text{B}}} = 1.4 \times {10^{ - 23}}\,{\text{J/K}}$$, $$\,{m_{{\text{He}}}} = 7 \times {10^{ - 27}}\,{\text{kg}}$$ and $$T = 300\,{\text{K}}$$ in the equation (4):

{V_{{\text{rms}}}} = \sqrt {\dfrac{{3{k_{\text{B}}}T}}{{{m_{{\text{He}}}}}}} \\
\Rightarrow{V_{{\text{rms}}}}= \sqrt {\dfrac{{3 \times 1.4 \times {{10}^{ - 23}},{\text{J/K}} \times 300,{\text{K}}}}{{7 \times {{10}^{ - 27}},{\text{kg}}}}} \\
\Rightarrow{V_{{\text{rms}}}}= \sqrt {1.8 \times {{10}^6}} \\
\therefore{V_{{\text{rms}}}}= 1.34 \times {10^3},{\text{m}}{{\text{s}}^{ - 1}} \\

The velocity is found out to be: $$1.34 \times {10^3}\,{\text{m}}{{\text{s}}^{ - 1}} \sim 1.3 \times {10^3}\,{\text{m}}{{\text{s}}^{ - 1}}$$ **Hence, the correct option is (B).** **Note:** In this problem we are asked to find the velocity closest to the thermal velocity of a Helium atom at room temperature. Generally, while solving this problem, students tend to use the temperature in degree centigrade, but using this it will affect the result. Always use the kelvin scale of temperature, even if the temperature is provided in degree centigrade.