if the total energy of an electron in a hydrogen atom in an excited state is -3.4eV then what is the de-broglie wavelength of the electron

atomic structure

• Sam
• 8493 Views
• 2 Answers

• 2pi r = n wavelength .; here, r( radius of 2nd orbital)= 0.529× n^2 ;; r= 0.529 × 4 angstrom. ;; wavelenght = (2 × 3.14× 2.116)/2 answer =6.64 angstrom.

  • Rajeshwari
  • 09 Jun
  • 1 Comment

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• 2×pie ×r = 2wavelength [ R= 0.529×4×10^-10 ] WAVELENGTH= 6.64×10^-10

  • Arnav
  • 09 Jun
  • 0 Comment

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