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Question

In a hydrogen atom, the radius of the first orbit is \( 5.3 \times 10^{-11} \, \text{m} \). What is the
circumference of the third orbit?

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Choose one · Correct answer highlighted

Explanation

**Energy in hydrogen** total E_n = -13.6/n² eV, kinetic K = +13.6/n² eV, potential U = -27.2/n² eV, ratio K/U = -1/2, total negative indicates bound, zero at ionization, negative total means electron bound, requires energy to free. For n=4 E=-0.85 eV, U=-1.7 eV, K=0.85 eV, potential energy twice total negative. r_n = n² r₁ , r₃ = 3² × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m . Circumference = 2π r₃ = 2 × 3.14 × 4.77 × 10⁻¹⁰ ≈ 3.0 × 10⁻⁹ m . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE

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