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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De Broglie wavelength associated with He atom = 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...
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De Broglie wavelength associated with He atom = 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 a gas at temperature T,is given as: De Broglie wavelength is given by the relation: Where, m = Mass of a He atom We have the ideal gas formula: PV = RT PV = kNT 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: Hence, the mean separation between the atoms is much greater than the de Broglie wavelength. read less
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