NEET 2025 · PhysicsAngular momentumPrevious Year Question
The de Broglie wavelength of an electron in the n = 2 state of hydrogen atom is close to: (Given Bohr radius = 0.052 nm)
- A.1.67 nm
- B.2.67 nm
- C.0.067 nm
- D.0.67 nm✓
Correct Answer
(D) 0.67 nm
Solution & Explanation
Answer: (d) 0.67 nm Solution: de Broglie wavelength in nth orbit: λₙ = 2πrₙ/n with rₙ = a₀n². For n=2: r₂ = 0.052×2² = 0.208 nm. λ₂ = 2π(0.208)/2 = π(0.208) = 0.653 nm ≈ 0.67 nm. (Equivalently λₙ = 2πa₀n, growing linearly with n.)
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