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Question
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)
Solution
Temperature of the nitrogen molecule, T = 300 K
Atomic mass of nitrogen = 14.0076 u
Hence, mass of the nitrogen molecule, m = 2 × 14.0076 = 28.0152 u
But 1 u = 1.66 × 10−27 kg
∴ m = 28.0152 × 1.66 × 10−27 kg
Planck’s constant, h = 6.63 × 10−34 Js
Boltzmann constant, k = 1.38 × 10−23 J K−1
We have the expression that relates mean kinetic energy `(3/2 "kT")` of the nitrogen molecule with the root mean square speed `("v"_("rms"))` as:
`1/2 "mv"_"rms"^2 = 3/2 "kT"`
`"v"_"rms" = sqrt((3"kT")/"m")`
Hence, the de Broglie wavelength of the nitrogen molecule is given as:
`lambda = "h"/("mv"_"rms") = "h"/sqrt(3 "mkT")`
`= (6.63 xx 10^(-34))/sqrt(3xx28.0152 xx 1.66 xx 10^(-27) xx 1.38 xx 20^(-23) xx 300 )`
= 0.028 × 10−9 m
= 0.028 nm
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.
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