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पाठ्यक्रम

De Broglie's Explanation

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Estimated time: 6 minutes
Maharashtra State Board: Class 12

Key Points: de Broglie's Explanation

de Broglie Wavelengths for Charged Particles (accelerated through potential V)

Particle Mass de Broglie Wavelength
Electron me = 9.1 × 10−31 kg \[\lambda=\frac{12.27}{\sqrt{V}}\] Å
Proton mp = 1.67 × 10−27 kg \[\lambda=\frac{0.286}{\sqrt{V}}\] Å
Deuteron md = 2 × 1.67 × 10−27 kg \[\lambda=\frac{0.202}{\sqrt{V}}\] Å
α-particle mα = 4 × 1.67 × 10−27 kg \[\lambda=\frac{0.101}{\sqrt{V}}\] Å

de Broglie Wavelengths for Uncharged Particles:

Particle/Condition Formula
Neutron \[\lambda=\frac{h}{\sqrt{2mK}}=\frac{6.62\times10^{-34}}{\sqrt{2\times1.67\times10^{-27}K}}\]
Thermal neutron (at temp T) \[\lambda=\frac{h}{\sqrt{2mkT}}=\frac{30.835}{\sqrt{T}}\] Å
Gas molecules at temp T \[\lambda=h/mv_{rms},\mathrm{energy~}K=\frac{3}{2}kT\to\lambda=\frac{h}{\sqrt{3mkT}}\]

Key Derivation Logic:

  • Planck's quantum theory: photon energy E = hν, de Broglie wavelength λ = h/p
  • If a photon has energy E = hν, treating it as mass m by relativity: E = mc2, so p = mc = h/λ
  • For a material particle: momentum p = mv, so de Broglie wavelength λ = h/(mv)
  • Kinetic energy \[K=p^2/2m\to\lambda=h/\sqrt{2mK}\]
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