The de Broglie wavelength associated with a nitrogen molecule at atmospheric pressure and temperature `27^@C` will be nearly

Options

A. 0.1 Å
B. 0.2 Å
C. 0.3 Å
D. 0.4 Å

Answer

To find the de Broglie wavelength associated with a nitrogen molecule (N₂) at atmospheric pressure and a temperature of 27°C, we will follow these steps:

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin**:
- The temperature in Celsius is given as 27°C.
- To convert to Kelvin, use the formula:
\[ T(K) = T(°C) + 273.15 \]
- Thus,
\[ T = 27 + 273.15 = 300.15 \approx 300 \, K \]

2. **Determine the Mass of a Nitrogen Molecule**:
- A nitrogen molecule (N₂) consists of two nitrogen atoms. The atomic mass of nitrogen (N) is approximately 14 u (atomic mass units).
- Therefore, the molecular mass of N₂ is:
\[ m = 2 \times 14 \, u = 28 \, u \]
- To convert atomic mass units to kilograms, use the conversion factor \(1 \, u = 1.66 \times 10^{-27} \, kg\):
\[ m = 28 \times 1.66 \times 10^{-27} \, kg = 4.648 \times 10^{-26} \, kg \]

3. **Calculate the Kinetic Energy of the Molecule**:
- The average kinetic energy (KE) of a molecule at temperature T is given by:
\[ KE = \frac{3}{2} k_B T \]
- Where \(k_B\) (Boltzmann's constant) is approximately \(1.38 \times 10^{-23} \, J/K\).
- Substituting the values:
\[ KE = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 300 \]
\[ KE \approx 6.21 \times 10^{-21} \, J \]

4. **Calculate the de Broglie Wavelength**:
- The de Broglie wavelength (\(\lambda\)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
- Where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, J \cdot s\)) and \(p\) is the momentum.
- The momentum \(p\) can be expressed in terms of kinetic energy:
\[ p = \sqrt{2m \cdot KE} \]
- Substituting the values:
\[ p = \sqrt{2 \times (4.648 \times 10^{-26}) \times (6.21 \times 10^{-21})} \]
\[ p \approx \sqrt{5.77 \times 10^{-46}} \approx 7.59 \times 10^{-23} \, kg \cdot m/s \]

5. **Final Calculation of Wavelength**:
- Now substituting \(p\) back into the wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{7.59 \times 10^{-23}} \approx 8.73 \times 10^{-12} \, m \]
- Converting meters to angstroms (1 angstrom = \(10^{-10} \, m\)):
\[ \lambda \approx 0.0873 \, nm = 0.873 \, \text{angstroms} \]

6. **Conclusion**:
- The de Broglie wavelength associated with a nitrogen molecule at atmospheric pressure and temperature of 27°C is approximately \(0.3 \, \text{angstroms}\).