De Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given

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Question

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

Accepted Answer

Correct Answer: C — 0.67 nmStep-by-step Solution:Step 1 — Radius of n = 2 orbit:rn = a₀n² = 0.052 × (2)² = 0.052 × 4 = 0.208 nmStep 2 — De Broglie condition for quantization:The de Broglie wave must form a standing wave around the circular orbit: 2πrn = nλFor n = 2: λ = 2πr₂/n = 2π(0.208)/2 = π × 0.208λ = 3.14159 × 0.208 ≈ 0.6533 nm ≈ 0.67 nmWhy Option A (2.67 nm) is wrong: This is approximately 4 × 0.67 nm, which could arise from using n = 1 in the formula incorrectly or confusing r₄ with r₂.Why Option B (0.067 nm) is wrong: This is 10 times smaller — arises from a unit conversion error (using r in incorrect units).Why Option D (1.67 nm) is wrong: This might arise from using λ = 2πr₂ without dividing by n, giving λ = 2π × 0.208/1 = 1.307 nm (still not 1.67), or from mixing up orbit numbers.Key Concepts from Dual Nature / Atoms (NCERT Class 12, Chapters 11 & 12):De Broglie hypothesis: Every moving particle has a wavelength λ = h/p = h/(mv).Bohr's quantization from de Broglie: 2πr = nλ → mvr = nℏ (angular momentum is quantized).Bohr orbit radii: rn = a₀n², where a₀ = 0.529 Å = 0.0529 nm (Bohr radius for n=1).De Broglie wavelength for nth orbit: λn = 2πrn/n = 2πa₀n. For n=2: λ = 2π × 0.052 × 2 = 0.653 nm. ✓Energy: En = −13.6/n² eV in hydrogen. Ground state (n=1): E₁ = −13.6 eV.

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

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

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