An electron revolves round Li^2+ nucleus at a distance of 1.587 Angstrom. The speed of electron shou — Atomic Structure Chemistry Question
Question
An electron revolves round Li^2+ nucleus at a distance of 1.587 Angstrom. The speed of electron should be
Answer: A
💡 Solution & Explanation
r = 0.529*n^2/Z => 1.587 = 0.529*n^2/3 => n = 3. v = 2.188*10^6*Z/n = 2.188*10^6*3/3 = 2.188*10^6 m/s.
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