Class 12 Physics
CBSE and UP Board
Dual Nature of Radiation and Matter
Chapter-11 Exercise 11.19
NCERT Solutions for Class 12 Physics Chapter 11 Question-19
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
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Temperature of the nitrogen molecule, T = 300 K
Atomic mass of nitrogen = 14.0076 u
Hence, mass of the nitrogen molecule, m = 2 x 14.0076 = 28.0152 u
But 1 u = 1.66 x 10⁻27 kg
Therefore, m = 28.0152 x 1.66 x 10⁻27 kg
Planck’s constant, h = 6.63 x 10⁻34 Js
Boltzmann constant, k = 1.38 x 10⁻23J K⁻1
We have the expression that relates mean kinetic energy (3KT/2 )of the nitrogen molecule with the root mean square speed (vrms) as:
1/2 x m (vrms)2 = 3/2 kT
vrms= √(3KT/m)
Hence, the de Broglie wavelength of the nitrogen molecule is given as:
λ = h/(mvrms)= h/√(3mKT)
=(6.63 x 10⁻34) /√[3 x 28.0152 x (1.66 x 10⁻27) x (1.38 x 10⁻23) x 300
= 0.028 x 10⁻⁹ m
= 0.028 nm
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.