Question:
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 degree C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.
Answer:
De broglie wavelength associated with He atom = 0.7268 x 10-10 m
Room temp. T = 27°C = 27 +273 K = 300 K
Atmospheric condition = 1 atm = 1.01 x 105 Pa
Atomic weight of He atom = 4 amu
Avogadros number = n = 6.023 x 1023
Boltzmann constant = k = 1.23 x 10-23 J mol-1 K-1
The average energy of a gas molecule at temperature T is given as E= (3/2) kT
De broglie wavelength = λ = h/√(2mE) [m = mass of He atom]
So, m = Atomic weight / n
= 4 / 6.023 x 1023
= 6.64 x 10-27 kg
According to the ideal gas equation, PV = RT
Or, PV = kNT
So, V/N = kT/P [V = vol of gas, N = no. of moles of gas]
Mean separation between two atoms of a gas is given by r = (V/N)1/3 = (kT/P)1/3
= 1.38 x 10-23 x 300 / 1.01 x 105
= 3.35 x 10-9 metres.
Thus, the mean separation between the atoms is much greater than the de broglie wavelength.
Not what you are looking for? Go ahead and submit the question, we will get back to you.
