What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean- square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

To find the de Broglie wavelength of a nitrogen molecule in air at 300 K, we first need to calculate the root-mean-square (rms) speed of the nitrogen molecules and then use that speed to determine the de Broglie wavelength. The de Broglie wavelength is given by the formula:

Calculating the Root-Mean-Square Speed

The root-mean-square speed ($v_{rms}$) of gas molecules can be calculated using the equation:

$v_{rms}=\sqrt{\frac{3kT}{m}}$

Where:

• k is the Boltzmann constant, approximately $1.38\times {10}^{-23}\text{J/K}$.
• T is the absolute temperature in Kelvin (300 K in this case).
• m is the mass of a single nitrogen molecule in kilograms.

First, we need to convert the atomic mass of nitrogen from atomic mass units (u) to kilograms. The atomic mass of nitrogen is given as 14.0076 u. Since 1 u is approximately $1.66\times {10}^{-27}\text{kg}$, we can calculate:

$m=14.0076\text{u}\times 1.66\times {10}^{-27}\text{kg/u}\approx 2.32\times {10}^{-26}\text{kg}$

Now, substituting the values into the $v_{rms}$ formula:

$v_{rms}=\sqrt{\frac{3\times (1.38\times {10}^{-23}\text{J/K})\times 300\text{K}}{2.32\times {10}^{-26}\text{kg}}}$

Calculating the numerator:

$3\times (1.38\times {10}^{-23})\times 300\approx 1.242\times {10}^{-20}\text{J}$

Now, dividing by the mass:

$v_{rms}=\sqrt{\frac{1.242\times {10}^{-20}}{2.32\times {10}^{-26}}}\approx \sqrt{5.35\times {10}^5}\approx 731.3\text{m/s}$

Finding the de Broglie Wavelength

Now that we have the rms speed, we can calculate the de Broglie wavelength ($\lambda$) using the formula:

$\lambda =\frac{h}{mv}$

Where:

• h is Planck's constant, approximately $6.626\times {10}^{-34}\text{J s}$.
• v is the speed of the molecule (which we found to be $v_{rms}$).

Substituting the values into the de Broglie wavelength formula:

$\lambda =\frac{6.626\times {10}^{-34}\text{J s}}{(2.32\times {10}^{-26}\text{kg})\times (731.3\text{m/s})}$

Calculating the denominator:

$2.32\times {10}^{-26}\times 731.3\approx 1.698\times {10}^{-23}\text{kg m/s}$

Now, we can find the wavelength:

$\lambda \approx \frac{6.626\times {10}^{-34}}{1.698\times {10}^{-23}}\approx 3.90\times {10}^{-11}\text{m}$

Final Result

Thus, the de Broglie wavelength of a nitrogen molecule in air at 300 K is approximately $3.90\times {10}^{-11}\text{m}$ or $39.0\text{pm}$ (picometers). This wavelength is extremely small, reflecting the quantum nature of particles at the molecular level.