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PhysicsNEET 2025Matter waves and de Broglie wavelength

Dual Nature of Radiation and Matter

The de Broglie wavelength of an electron orbiting in the n=2n = 2 state of the hydrogen atom is close to (given Bohr radius = 0.052 nm):

  1. A.0.067 nm
  2. B.0.67 nm✓
  3. C.1.67 nm
  4. D.2.67 nm

Solution

In the nnth orbit the circumference equals nn de Broglie wavelengths: 2πrn=nλ2\pi r_n = n\lambda. With r2=4a0=0.208r_2 = 4a_0 = 0.208 nm: λ=2π×0.2082≈0.65\lambda = \frac{2\pi \times 0.208}{2} \approx 0.65 nm ≈0.67\approx 0.67 nm.

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