The de-Broglie wavelength of a molecule of thermal energy k B T (k B = Boltzmann constant…

Physics · JEE Main · NTA ExamsAtoms and Nuclei

The de-Broglie wavelength of a molecule of thermal energy kBT (kB = Boltzmann constant and T = absolute temperature), is
  1. \(\lambda=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mk}_{\mathrm{B}} \mathrm{~T}}}\)
  2. \(\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}_{\mathrm{B}} \mathrm{~T}}}\)
  3. \(\mathrm{h} \sqrt{2 \mathrm{mk}_{\mathrm{B}} \mathrm{~T}}\)
  4. \(\frac{\mathrm{h}}{4 \mathrm{~m}^{2} \mathrm{k}_{\mathrm{B}}^{2} \mathrm{~T}^{2}}\)

Answer

(B) h 2 mk _ B ~T

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