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

\frac{h}{\sqrt{2 m K T}}

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\frac{h}{2 m K T}

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h \sqrt{2 m K T}

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\frac{1}{h \sqrt{2 m K T}}

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Solution

The correct option is A $\frac{h}{\sqrt{2 m K T}}$
$\frac{1}{2} m v^{2} = K T$
$m v = \sqrt{2 m K T}$
So de-broglie wavelength
$λ = \frac{h}{m v}$
$\to λ = \frac{h}{\sqrt{2 m K T}}$

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