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  4. At temperature T, the average kinetic energy of any particle is (3/2) kT. The de Broglie wavelength follows the order :

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

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Solution:

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

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

$\frac{1}{2}mv^{2}=\frac{3}{2}KT$

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

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

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)