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Question

What is the speed of an electron in the \( n = 5 \) orbit of a hydrogen atom if its speed in \( n = 1
\) is \( 2.2 \times 10^6 \, \text{m/s} \)?

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Explanation

**Bohr's quantization** angular momentum L = m v r = n h/2π, for n=2 L=2h/2π= h/π=2.11×10⁻³⁴ J·s, for n=5 L=5h/2π, de Broglie λ = h/p, p= m v, for first orbit v=2.2×10⁶ m/s, λ= h/(m v)=6.6×10⁻³⁴/(9.1×10⁻³¹×2.2×10⁶)=3.3×10⁻¹⁰ m, circumference 2πr=3.33×10⁻¹⁰ m, one wavelength fits for n=1. v_n = (v₁/n) . For n = 5 : v₅ = (2.2 × 10⁶/5) = 4.4 × 10⁵ m/s . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 4.4 × 10⁵ m/s, consistent with Bohr model and nuclear binding energy systematics.

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