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For the n = 2 stationary orbit of a hydrogen atom (Bohr radius a0 =…
Question
For the n = 2 stationary orbit of a hydrogen atom (Bohr radius a0 = 0.529 angstrom), the de Broglie wavelength of the orbiting electron is exactly an integer fraction of the orbit circumference. Compute that de Broglie wavelength in angstrom.
✓ Verified answer: 6.6476checked by our engine — not a guess
Step-by-step solution
Bohr quantization: 2*pi*r_n = n*lambda, so lambda = 2*pi*r_n/n.
r_n = n^2 a0, so r2 = 4*0.529 = 2.116 angstrom.
lambda = 2*pi*2.116/2 = pi*2.116 = 6.6476 angstrom.
Final answer6.6476
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