The de-Broglie wavelength of a molecule of thermal energy k B T (k B = Boltzmann constant…
Physics · JEE Main · NTA Exams — Atoms and Nuclei
The de-Broglie wavelength of a molecule of thermal energy kBT (kB = Boltzmann constant and T = absolute temperature), is
- \(\lambda=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mk}_{\mathrm{B}} \mathrm{~T}}}\)
- \(\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}_{\mathrm{B}} \mathrm{~T}}}\)
- \(\mathrm{h} \sqrt{2 \mathrm{mk}_{\mathrm{B}} \mathrm{~T}}\)
- \(\frac{\mathrm{h}}{4 \mathrm{~m}^{2} \mathrm{k}_{\mathrm{B}}^{2} \mathrm{~T}^{2}}\)
Answer
(B) h 2 mk _ B ~T
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