Question

At temperature T, the average kinetic energy of any particle is $\frac{3}{2}$ kT. The de Broglie wavelength follows the order :

• A. Thermal proton > Visible photon > Thermal electron
• B. Thermal proton > Thermal electron > Visible photon
• C. Visible photon > Thermal electron > Thermal neutron
• D. Visible photon > Thermal neutron > Thermal electron

Answer 1

Answer: (C)

Kinetic energy of any particle $=\frac{3}{2}KT$

Also $K.E.=\frac{1}{2}m{v}^2$

$\frac{1}{2}m{v}^2=\frac{3}{2}KT$

$\Rightarrow v^2=\frac{3KT}{m}$

$v=\sqrt{\frac{3KT}{m}}$

De-broglie wavelength

$=\lambda =\frac{h}{mv}=\frac{h}{m\sqrt{\frac{3KT}{m}}}$

$\lambda =\frac{h}{\sqrt{3KTm}}\lambda \propto \frac{1}{\sqrt{m}}$

Mass of electron < mass of neutron $\lambda$(electron) > $\lambda$ (neutron)