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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 19 of 37
Marks: +1, -0
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, K = 1.38 × 10−23 J K−110^{-23}\,\mathrm{J}\,\mathrm{K}^{-1}).
Solution:  
Let us first calculate mass of each N2 molecule.
m = 2 × 14. 0076 × 1.66 × 10−2710^{-27} kg = 46.5 × 10−2710^{-27} kg
T = 300 K
Average kinetic energy of N2\mathrm{N}_2 molecules at temperature T
12mvmax⁡2\frac{1}{2} m v_{\max}^2 = 32\frac{3}{2} kT or vrmsv_{\mathrm{rms}} = 3kTm\sqrt{\frac{3kT}{m}}
∴ λ = hmvrms\frac{h}{m v_{\mathrm{rms}}} = h3mkT\frac{h}{\sqrt{3mkT}}
=
6.63×10−343×46.5×10−27×1.38×10−23×300\frac{6.63 \times 10^{-34}}{\sqrt{3 \times 46.5 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}}
m
= 6.63×10−34577.53×10−24\frac{6.63 \times 10^{-34}}{\sqrt{577.53} \times 10^{-24}} = 6.63×10−1024.03\frac{6.63 \times 10^{-10}}{24.03} m = 0.0276 × 10−1910^{-19} m = 0.028 nm
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