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Step-by-Step Solution
Step 1: Write down the formula for RMS speed
For an ideal gas, the root-mean-square (RMS) speed of a molecule/atom is given by:
$v_{rms} = \sqrt{\frac{3 k_B T}{m}}$
Step 2: Identify the given data
• Boltzmann constant, $k_B = 1.4 \times 10^{-23}\,\text{J/K}$
• Temperature, $T = 300\,\text{K}$
• Mass of Helium atom, $m = 7 \times 10^{-27}\,\text{kg}$
Step 3: Substitute the values into the formula
$v_{rms} = \sqrt{\frac{3 \times (1.4 \times 10^{-23}) \times 300}{7 \times 10^{-27}}}
$
Step 4: Simplify the expression
First, simplify the numerator and denominator inside the square root:
$\displaystyle v_{rms} = \sqrt{\frac{3 \times 1.4 \times 300 \times 10^{-23}}{7 \times 10^{-27}}}
= \sqrt{\frac{3 \times 1.4 \times 300}{7} \times \frac{10^{-23}}{10^{-27}}}
= \sqrt{\frac{3 \times 1.4 \times 300}{7} \times 10^{4}}.
$
Now, compute step by step:
The factor $\frac{3 \times 1.4 \times 300}{7}$ simplifies to approximately $3 \times (1.4 \times 300 / 7)$.
$1.4 \times 300 = 420$, and $420 / 7 = 60$.
So inside the square root we have $3 \times 60 \times 10^{4} = 180 \times 10^{4}.$
$\displaystyle v_{rms} = \sqrt{180 \times 10^{4}}
= \sqrt{1.8 \times 10^{6}}
= \sqrt{1.8} \times 10^{3}.
$
Since $\sqrt{1.8} \approx 1.34$, we get:
$v_{rms} \approx 1.34 \times 10^{3}\, \text{m/s}.$
Step 5: State the final result
The RMS speed (thermal velocity) of a Helium atom at room temperature ($300\,\text{K}$) is approximately $1.3 \times 10^{3}\,\text{m/s}$.
