At temperature T, the average kinetic energy of any particle is 3/2 KT. The de Broglie wavelength follows the order :
(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
Correct answer: (d) Visible photon > Thermal electron > Thermal neutron
Explanation:
Kinetic energy of any particle = 3/2 KT
Also K.E = \(\frac{1}{2}mv^2\)
\(\frac{1}{2}mv^2\) = \(\frac{3}{2}\)KT
⇒ \(v^2 = \frac{3KT}{m}\)
v = \(\sqrt{\frac{3KT}{m}}\)
De-broglie wavelength
= λ = \(\frac{h}{mv}\) = \(\frac{h}{m\sqrt{\frac{3KT}{m}}}\)
λ = \(\frac{h}{\sqrt{3KTm}}\) λ ∝ \(\frac{1}{\sqrt m}\)
Mass of electron < mass of neutron λ (electron) > λ (neutron)