Question

de-Broglie wavelength of atom at $TK$ absolute temperature will be

• A. $\frac{h}{mKT}$
• B. $\frac{h}{\sqrt{3mKT}}$
• C. $\frac{\sqrt{2mKT}}{h}$
• D. $\sqrt{2mKT}$

Answer 1

The correct option is B $\frac{h}{\sqrt{3mKT}}$

de-Broglie wavelength $\lambda =\frac{h}{mv}$
Kinetic energy of the atom $K.E=\frac{1}{2}m{v}^2=\frac{3}{2}KT$
$⟹mv=\sqrt{3mKT}$
Thus we get $\lambda =\frac{h}{\sqrt{3mKT}}$
So, option B is correct.