Class 12 Physics
CBSE and UP Board
Dual Nature of Radiation and Matter
Chapter-11 Exercise 11.35
NCERT Solutions for Class 12 Physics Chapter 11 Question-35
Additional Exercise
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 ºC) 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 = 0.7268 x 10-10 m
Room temperature, T = 27°C = 27 + 273 = 300 K
Atmospheric pressure, P = 1 atm = 1.01 x 10⁵ Pa
Atomic weight of a He atom = 4
Avogadro’s number, Na = 6.023 x 1023
Boltzmann constant, k= 1.38 x 10-23 J mol-1 K-1
Average energy of a gas at temperature T, is given as:
E= 3/2 kT
De Broglie wavelength is given by the relation:
λ = h/√[2mE]
Where,
m = Mass of a He atom
= Atomic weight /Na = 4/(6.023 x 1023)
=6.64 x 10⁻²⁴ g = 6.64 x 10⁻²⁷
Therefore , λ = h/√[3mkT]
= (6.6 x 10⁻34) /√[3 x (6.64 x 10⁻²⁷) x (1.38 x 10-23) x 300]
= 0.7268 x 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:
r³ = (V/N) = (kT/P) = [(1.38 x 10-23) x (300) /( 1.01 x 10⁵)]
r = 3.35 x 10⁻⁹ m
Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.