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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 35 of 37
Marks: +1, -0
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.
Solution:  
Let us first find mass ‘m’ of each helium atom.
m = Atomic weight of HeliumAvogadro’s number\frac{\text{Atomic weight of Helium}}{\text{Avogadro's number}}
m = 46×1023\frac{4}{6 \times 10^{23}} g = 23×10−23\frac{2}{3} \times 10^{-23} g = 23×10−26\frac{2}{3} \times 10^{-26} kg
Absolute temperature, T = 273 + 27 = 300 K
Average K.E. of a He atom at absolute temperature T
K = 12mv2\frac{1}{2} m v^{2} = 32\frac{3}{2} kT
m2v2m^{2} v^{2} = p2p^{2} = 3 mkT
Momentum, p = 3mkT\sqrt{3 m k T}
Wavelength of the wave associated with He atom at room temperature
λ = hp\frac{h}{p} = h3mkT\frac{h}{\sqrt{3 m k T}} =
6.63×10−343×23×10−26×1.38×10−23×300\frac{6.63 \times 10^{-34}}{\sqrt{3 \times \frac{2}{3} \times 10^{-26} \times 1.38 \times 10^{-23} \times 300}}
λ = 0.73 × 10−1010^{-10} m ... (i)
Now let us find mean separation between He atoms.
Mean separation, r = [Molar volumeAvagadro’s number]13\left[ \frac{\text{Molar volume}}{\text{Avagadro's number}} \right]^{\frac{1}{3}}
Here for 1 mole, PV = RT
PV = NkT or VN\frac{V}{N} = kTP\frac{k T}{P}
So, mean separation r = [kTP]13\left[ \frac{k T}{P} \right]^{\frac{1}{3}}
Here, k = Boltzman’s constant = 1.38 × 10−23JK−110^{-23} \mathrm{J} \mathrm{K}^{-1}
T = Absolute temperature = 300 K
P = Atmospheric pressure = 1.01 × 10510^{5} Pa
r = [1.38×10−23×3001.01×1058]13\left[ \frac{1.38 \times 10^{-23} \times 300}{1.01 \times 10^{58}} \right]^{\frac{1}{3}} m = 3.4 × 10−910^{-9} m ... (ii)
Comparing the wavelength ‘λ’ with mean separation ‘r’, it can be observed that separation is larger than wavelength, i.e. (r ≫ λ)
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