Question

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

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

Answer 1

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

Key Concept The average kinetic energy of atom at temperature $TK$ is
$E=\frac{3}{2}KT$
Now, kinetic energy of an atom $E=\frac{P^2}{2m}$
$\therefore P=\sqrt{2ME}$
$\therefore$ de-Broglie wavelength
$\lambda =\frac{h}{P}=\frac{h}{\sqrt{2mE}}$
$\Rightarrow \lambda =\frac{h}{\sqrt{2m\times \frac{3}{2}KT}}=\frac{h}{\sqrt{3mKT}}$.