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
Solution :
[d] 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)
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