What is the de-Broglie wavelength of nitrogen molecule in air at 300K ? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of nitrogen `=14.0076U`) Plank's constant=`6.63xx10^(-34)Js` , Boltzmann constant `=1.38xx10^(-23)JK^(-1)`

Options

A. `2.75xx10^(-11)m`
B. `2.75xx10^(-12)m`
C. `3.24xx10^(-11)m`
D. `3.24xx10^(-12)m`

Explanation

To find the de-Broglie wavelength of a nitrogen molecule in air at 300K, we will follow these steps:

### Step 1: Calculate the root mean square speed (Vrms) of the nitrogen molecule.

The formula for the root mean square speed is given by:

\[ V_{rms} = \sqrt{\frac{3kT}{m}} \]

Where:
- \( k \) is the Boltzmann constant \( = 1.38 \times 10^{-23} \, J/K \)
- \( T \) is the temperature in Kelvin \( = 300 \, K \)
- \( m \) is the mass of the nitrogen molecule.

### Step 2: Calculate the mass of the nitrogen molecule.

Since nitrogen exists as a diatomic molecule (N₂), the mass of one nitrogen molecule is:

\[ m = 2 \times \text{atomic mass of nitrogen} = 2 \times 14.0076 \, U \]

To convert atomic mass units (U) to kilograms, we use the conversion factor \( 1 \, U = 1.66 \times 10^{-27} \, kg \):

\[ m = 2 \times 14.0076 \, U \times 1.66 \times 10^{-27} \, kg/U \]

Calculating this gives:

\[ m = 28.0152 \, U \times 1.66 \times 10^{-27} \, kg/U = 4.653 \times 10^{-26} \, kg \]

### Step 3: Substitute the values into the Vrms formula.

Now substituting the values of \( k \), \( T \), and \( m \) into the Vrms formula:

\[ V_{rms} = \sqrt{\frac{3 \times (1.38 \times 10^{-23}) \times 300}{4.653 \times 10^{-26}}} \]

Calculating this gives:

\[ V_{rms} = \sqrt{\frac{1.242 \times 10^{-20}}{4.653 \times 10^{-26}}} \approx \sqrt{2.67 \times 10^{5}} \approx 517.3 \, m/s \]

### Step 4: Calculate the de-Broglie wavelength.

The de-Broglie wavelength \( \lambda \) is given by:

\[ \lambda = \frac{h}{p} \]

Where \( p \) is the momentum given by \( p = mv \). Since we are using the root mean square speed, we have:

\[ \lambda = \frac{h}{m \cdot V_{rms}} \]

Substituting the values:

\[ \lambda = \frac{6.63 \times 10^{-34}}{4.653 \times 10^{-26} \cdot 517.3} \]

Calculating this gives:

\[ \lambda \approx \frac{6.63 \times 10^{-34}}{2.41 \times 10^{-43}} \approx 2.75 \times 10^{-11} \, m \]

### Final Answer:
The de-Broglie wavelength of a nitrogen molecule in air at 300K is approximately:

\[ \lambda \approx 2.75 \times 10^{-11} \, m \]

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