Atomic StructureEasy

Question

An electron revolves round Li2+ nucleus at a distance of 1.587 Å. The speed of electron should be

Options

A.2.188 × 106 m/s
B.6.564 × 106 m/s
C.7.293 × 105 m/s
D.7.293 × 106 m/s

Solution

$r_{n,z} = 0.529 \times \frac{n^{2}}{z} \Rightarrow 1.587 = 0.529 \times \frac{n^{2}}{3} \Rightarrow n = 3$

$\therefore V_{n,z} = 2.188 \times 10^{6} \times \frac{z}{n} = 2.188 \times 10^{6} \times \frac{3}{3} = 2.188 \times 10^{6}\text{ m/s}$

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