Q1.De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm)
- A0.067 nm
- B0.67 nmAnswer
- C1.67 nm
- D2.67 nm
Bohr quantisation makes the orbit circumference an exact whole number of de Broglie wavelengths: 2πrn = nλ. With r₂ = 0.052 x 2² = 0.208 nm, this gives λ = 2πr₂/2 = π x 0.208 ≈ 0.65 nm.
Source: NEET 2025 (4 May), Code 45, Q34