JEE Mains · Chemistry · STD 11 - 2. structure of atom
At temperature \(T\), the average kinetic energy of any particle is \(\frac{3}{2} KT\) . 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
Answer & Solution
Correct Answer
(D) Visible photon \(>\) Thermal electron \(>\) Thermal neutron
Step-by-step Solution
Detailed explanation
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…
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