Atomic Structure Question 396

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

Options:

A) Visible photon > Thermal neutron > Thermal electron

B) Thermal proton > Thermal electron > Visible photon

C) Thermal proton > Visible photon > Thermal electron

D) Visible photon > Thermal electron > Thermal neutron

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Answer:

Correct Answer: D

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)



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