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Question: What is the de Broglie wavelength of a nitrogen molecule in air at \(300{\text{ K}}\)? Assume that t...

What is the de Broglie wavelength of a nitrogen molecule in air at 300 K300{\text{ 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 = 14.0076{\text{ u}})

Explanation

Solution

To solve this we must know the expression for the mean kinetic energy of gas molecules and the expression for the kinetic energy of gas molecules. The kinetic theory of gases gives these two expressions. From these expressions derive the expression for de Broglie wavelength of a nitrogen molecule moving with the root-mean square speed.

Complete step-by-step answer:
We are given that the atomic mass of nitrogen is 14.0076 u14.0076{\text{ u}}. The molecular mass of nitrogen i.e. N2{{\text{N}}_{\text{2}}} is 28.0152 u28.0152{\text{ u}}. Where u=1.66×10−27 kg{\text{u}} = 1.66 \times {10^{ - 27}}{\text{ kg}}. Thus, molecular mass of nitrogen i.e. N2{{\text{N}}_{\text{2}}} is 28.0152×1.66×10−27 kg28.0152 \times 1.66 \times {10^{ - 27}}{\text{ kg}}.
We are given that the molecule is moving with the root-mean square speed of molecules at this temperature. Thus,
We know the expression for the mean kinetic energy of gas molecules is as follows:
KE=32kTKE = \dfrac{3}{2}kT …… (1)
Where KEKE is the mean kinetic energy,
kk is the Boltzmann constant,
TT is the temperature.
We know the expression for the kinetic energy of gas molecules is as follows:
KE=12mvrms2KE = \dfrac{1}{2}mv_{{\text{rms}}}^2 …… (2)
Where KEKE is the kinetic energy,
mm is the mass of the gas molecules,
vrms{v_{{\text{rms}}}} is the root mean square speed of the gas molecules.
Thus, from equation (1) and equation (2),
12mvrms2=32kT\dfrac{1}{2}mv_{{\text{rms}}}^2 = \dfrac{3}{2}kT
vrms2=32kT12mv_{{\text{rms}}}^2 = \dfrac{{\dfrac{3}{2}kT}}{{\dfrac{1}{2}m}}
vrms=3kTm{v_{{\text{rms}}}} = \sqrt {\dfrac{{3kT}}{m}} …… (3)
We know the expression for the de Broglie wavelength of a molecule is as follows:
λ=hmvrms\lambda = \dfrac{h}{{m{v_{{\text{rms}}}}}} …… (4)
Where, λ\lambda is the wavelength,
hh is the Planck’s constant,
mm is the mass of the gas molecules,
vrms{v_{{\text{rms}}}} is the root mean square speed of the gas molecules.
Thus, from equation (1) and equation (2),
λ=hm×3kTm\lambda = \dfrac{h}{{m \times \sqrt {\dfrac{{3kT}}{m}} }}
λ=h3kTm\lambda = \dfrac{h}{{\sqrt {3kTm} }}
Substitute 6.626×10−34 J s6.626 \times {10^{ - 34}}{\text{ J s}} for the Planck’s constant, 1.38×10−23 m2 kg s−2 K−11.38 \times {10^{ - 23}}{\text{ }}{{\text{m}}^2}{\text{ kg }}{{\text{s}}^{ - 2}}{\text{ }}{{\text{K}}^{ - 1}} for the Boltzmann constant, 300 K300{\text{ K}} for the temperature, 28.0152×1.66×10−27 kg28.0152 \times 1.66 \times {10^{ - 27}}{\text{ kg}} for the mass of gas. Thus,
λ=6.626×10−34 J s3×1.38×10−23 m2 kg s−2 K−1×300 K×28.0152×1.66×10−27 kg\lambda = \dfrac{{6.626 \times {{10}^{ - 34}}{\text{ J s}}}}{{\sqrt {3 \times 1.38 \times {{10}^{ - 23}}{\text{ }}{{\text{m}}^2}{\text{ kg }}{{\text{s}}^{ - 2}}{\text{ }}{{\text{K}}^{ - 1}} \times 300{\text{ K}} \times 28.0152 \times 1.66 \times {{10}^{ - 27}}{\text{ kg}}} }}
λ=0.275×10−10 m\lambda = 0.275 \times {10^{ - 10}}{\text{ m}}
Thus, the de Broglie wavelength of a nitrogen molecule in air is 0.275×10−10 m0.275 \times {10^{ - 10}}{\text{ m}}.

Note: Remember that the unit of atomic mass is u i.e. unified atomic mass unit. It is the one half of the mass of an unbound neutral atom of carbon-12 in its ground state. One unified atomic mass unit is equal to 1.66×10−27 kg1.66 \times {10^{ - 27}}{\text{ kg}}. Convert the atomic mass from a unified atomic mass unit to kilograms.