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

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Given that
Temperature T = 300 K
Atomic mass of nitrogen = 14.0076 amu
The molecular mass is the sum of the atomic masses of the two nitrogen atoms.
Nitrogen molecule mass m = 14.007 + 14.007 amu
m = 28.014 * 1.66 * 10^{-27} kg \because \text{ 1 amu } = 1,66 * 10^{-27} kg
Step 1: Expression for speed of molecules (root mean square speed)
Required formulas:
Kinetic energy K.E = \frac{1}{2} m(v_{rms})^2 ........................(1)
Where, m - mass and v_{rms} - root mean square speed
Gaseous particle(nitrogen) kinetic energy at a certain temperature K.E = \frac{3}{2} kT ................................(2)
Where, K - Boltzmann constant 1.38*10^{-23} J/K and T - temperature
From equation (1) and (2) we can write as follows
\frac{1}{2} m(v_{rms})^2 = \frac{3}{2} kT
(v_{rms})^2 = \frac{3kT}{m}
v_{rms}=\sqrt{\frac{3kT}{m}} ..........................(3)
Step 2: Equation for nitrogen de Broglie wavelength
de Broglie wavelength \lambda=\frac{h}{p}=\ \frac{h}{mv_{rms}}\ \ \ \ \ \ \ \ \ \ \
Where, h - Planck constant = 6.63* 10^{-34} J , p - linear momentum, m - mass and v_{rms} - speed
\lambda = \frac{h}{m \sqrt{\frac{3kT}{m}}} \because \text{ Equation (3)}
\lambda = \frac{h}{ \sqrt{3kTm}}
Step 3: Plug in the known value in the above equation
\lambda = \frac{6.63*10^{-34}}{3 * 1.38*10^{-23} * 300 * 28.014 * 1.66 * 10^{-27} }
\lambda = 0.028 * 10^{-9} m
\lambda = 0.028 nm
Therefore, de Broglie wavelength of nitrogen molecule \lambda = 0.028 nm