Calculating Helium Atom Thermal Velocity

This solution calculates the approximate thermal velocity of a Helium atom at room temperature using the provided constants.

Thermal Velocity Formula

The root-mean-square (RMS) thermal velocity ($v_{rms}$) of gas molecules is given by the formula derived from kinetic theory:

$v_{rms} = \sqrt{\frac{3k_B T}{m}}$

Where:

• $k_B$ is the Boltzmann constant
• $T$ is the absolute temperature in Kelvin
• $m$ is the mass of one molecule

Applying Given Values

We are given:

• Temperature, $T = 300$ K
• Boltzmann constant, $k_B = 1.4 \times 10^{-23}$ J/K
• Mass of Helium atom, $m_{He} = 7 \times 10^{-27}$ kg

Step-by-Step Calculation

1. Substitute the values into the formula:

   $v_{rms} = \sqrt{\frac{3 \times (1.4 \times 10^{-23} \text{ J/K}) \times (300 \text{ K})}{7 \times 10^{-27} \text{ kg}}}$

2. Simplify the numerator:

   $3 \times 1.4 \times 300 = 1260$

   So, the expression becomes:

   $v_{rms} = \sqrt{\frac{1260 \times 10^{-23}}{7 \times 10^{-27}}}$

3. Divide the numbers and combine the powers of 10:

   $\frac{1260}{7} = 180$

   $10^{-23} / 10^{-27} = 10^{-23 - (-27)} = 10^{4}$

   $v_{rms} = \sqrt{180 \times 10^{4}}$

4. Calculate the square root:

   $v_{rms} = \sqrt{180} \times \sqrt{10^{4}}$

   $v_{rms} = \sqrt{180} \times 10^{2}$

   Since $\sqrt{180}$ is approximately $13.416$, we get:

   $v_{rms} \approx 13.416 \times 10^{2} \text{ ms}^{-1}$

5. Express the result in the required format:

   $v_{rms} \approx 1.3416 \times 10^3 \text{ ms}^{-1}$

Conclusion on Thermal Velocity

The calculated value $1.3416 \times 10^3$ $ms^{-1}$ is closest to $1.3 \times 10^3$ $ms^{-1}$.