Question

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

• A. $\frac{h}{\sqrt{2mKT}}$
• B. $\frac{h}{2mKT}$
• C. $h\sqrt{2mKT}$
• D. $\frac{h}{h\sqrt{2mKT}}$

Answer 1

Answer: (A)

$\frac{h}{\sqrt{2mKT}}$

Answer 2

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