At temperature $\mathrm{T}$, the average kinetic energy of any particle is $\frac{3}{2} \mathrm{KT}$. The de…

At temperature $\mathrm{T}$, the average kinetic energy of any particle is $\frac{3}{2} \mathrm{KT}$. The de Broglie wavelength follows the order:
  1. Visible photon $>$ Thermal neutron > Thermal electron
  2. Thermal proton $>$ Thermal electron Visible photon
  3. Thermal proton $>$ Visible photon $>$ Thermal electron
  4. Visible photon $>$ Thermal electron Thermal neutron

Solution

Kinetic energy of any particle $=\frac{3}{2} \mathrm{KT}$
Also K.E. $=\frac{1}{2} \mathrm{mv}^{2}$
$\frac{1}{2} m v^{2}=\frac{3}{2} K T \Rightarrow v^{2}=\frac{3 K T}{m}$
$\mathrm{v}=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}}$
De-broglie wavelength $=\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{m} \sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}}}$
$\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{~K} \mathrm{Tm}}} \quad \lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$
Mass of electron $ < $ mass of neutron $\lambda$ (electron) $>\lambda$ (neutron)

Asked in: JEE-TOPICTESTS-CHEMISTRY

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