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Question
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27°C) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.
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Solution
De Broglie wavelength associated with He atom = 0.7268 × 10−10 m
Room temperature, T = 27°C = 27 + 273 = 300 K
Atmospheric pressure, P = 1 atm = 1.01 × 105 Pa
Atomic weight of a He atom = 4
Avogadro’s number, NA = 6.023 × 1023
Boltzmann constant, k = 1.38 × 10−23 J mol−1 K−1
Average energy of gas at temperature T, is given as:
`"E" = 3/2 "kT"`
De Broglie wavelength is given by the relation:
`lambda = "h"/(sqrt(2"mE"))`
Where,
m = Mass of a He atom
= `"Atomic weight"/"N"_"A"`
= `4/(6.023 xx 10^23)`
= 6.64 × 10−24 g
= 6.64 × 10−27 kg
∴ `lambda = "h"/sqrt(3"mkT")`
= `(6.6 xx 10^(-34))/sqrt(3xx6.64 xx10^(-27)xx 1.38 xx 10^(-23) xx 300)`
= 0.7268 × 10−10 m
We have the ideal gas formula:
PV = RT
PV = kNT
`"V"/"N" = "kT"/"P'`
Where,
V = Volume of the gas
N = Number of moles of the gas
Mean separation between two atoms of the gas is given by the relation:
`"r" = ("V"/"N")^(1/3) = ("kT"/"P")^(1/3)`
= `[(1.38 xx 10^(-23) xx 300)/(1.01 xx 10^5)]^(1/3)`
= 3.35 × 10−9 m
Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.
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