What is de-Broglie wavelength of a He-atom in a container at room temperature. `("use"U_(avg)=sqrt((8kT)/(pim)))`

Explanation

To find the de-Broglie wavelength of a helium atom in a container at room temperature, we can follow these steps:

### Step 1: Calculate the average speed (u_avg) of the helium atom
We use the formula for average speed:
\[ u_{avg} = \sqrt{\frac{8kT}{\pi m}} \]
Where:
- \( k \) is the Boltzmann constant, \( k = 1.38 \times 10^{-23} \, \text{J/K} \)
- \( T \) is the temperature in Kelvin, \( T = 298 \, \text{K} \)
- \( m \) is the mass of the helium atom in kilograms.

The molar mass of helium is approximately \( 4 \, \text{g/mol} \). To convert this to kilograms:
\[ m = \frac{4 \, \text{g/mol}}{1000 \, \text{g/kg}} \times \frac{1 \, \text{kg}}{6.022 \times 10^{23} \, \text{atoms/mol}} \approx 6.64 \times 10^{-27} \, \text{kg} \]

Now substituting the values into the equation:
\[ u_{avg} = \sqrt{\frac{8 \times (1.38 \times 10^{-23}) \times 298}{\pi \times (6.64 \times 10^{-27})}} \]

### Step 2: Calculate the value
Calculating the numerator:
\[ 8 \times (1.38 \times 10^{-23}) \times 298 \approx 3.28 \times 10^{-20} \]
Calculating the denominator:
\[ \pi \times (6.64 \times 10^{-27}) \approx 2.09 \times 10^{-26} \]
Now substituting these values:
\[ u_{avg} = \sqrt{\frac{3.28 \times 10^{-20}}{2.09 \times 10^{-26}}} \approx \sqrt{1.57 \times 10^{6}} \approx 1.25 \times 10^{3} \, \text{m/s} \]

### Step 3: Calculate the de-Broglie wavelength (λ)
The de-Broglie wavelength is given by the formula:
\[ \lambda = \frac{h}{mv} \]
Where:
- \( h \) is Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( m \) is the mass of the helium atom (calculated earlier)
- \( v \) is the average speed (calculated earlier)

Substituting the values:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(6.64 \times 10^{-27}) \times (1.25 \times 10^{3})} \]

### Step 4: Calculate the value of λ
Calculating the denominator:
\[ (6.64 \times 10^{-27}) \times (1.25 \times 10^{3}) \approx 8.30 \times 10^{-24} \]
Now substituting this into the wavelength equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{8.30 \times 10^{-24}} \approx 0.798 \times 10^{-10} \, \text{m} = 0.798 \, \text{Å} \]

### Final Answer
The de-Broglie wavelength of a helium atom at room temperature is approximately:
\[ \lambda \approx 0.79 \, \text{Å} \]