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)

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Solution

Mass of $N_{2} = 2 \times 14.0076 u = 28.0152 \times 1.66 \times 10^{- 27} k g$
$K E = 3 k T / 2 = m v_{r m s}^{2} / 2$

$\Rightarrow v_{r m s} = \sqrt{3 k T / m}$
de-broglie wavelength is, $h / m v_{r m s} = 0.028 n m$ .

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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)

Mass of N2=2×14.0076u=28.0152×1.66×10−27kgKE=3kT/2=mv2rms/2⇒vrms=√3kT/mde-broglie wavelength is, h/mvrms=0.028nm.

Mass of $N_{2} = 2 \times 14.0076 u = 28.0152 \times 1.66 \times 10^{- 27} k g$
$K E = 3 k T / 2 = m v_{r m s}^{2} / 2$

$\Rightarrow v_{r m s} = \sqrt{3 k T / m}$
de-broglie wavelength is, $h / m v_{r m s} = 0.028 n m$ .