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  4. The de Broglie Wavelength of a molecule of thermal energy KT (K is Boltzmann constant. and T is absolute temperature) is given by

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Q. The de Broglie Wavelength of a molecule of thermal energy KT (K is Boltzmann constant. and T is absolute temperature) is given by

Dual Nature of Radiation and Matter

Solution:

$\frac{1}{2} mv ^{2}= KT$
$mv =\sqrt{2 mKT }$
So de-broglie wavelength
$\lambda=\frac{ h }{ mv }$
$\rightarrow \lambda=\frac{ h }{\sqrt{2 mKT }}$