Atomic StructureEasy

Question

The speed of electron revolving in the fourth orbit of a hydrogen-like atom or ion is 1094 km/s. The atom or ion is

Options

A.H
B.He+
C.Li2+
D.Be3+

Solution

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

Or, $1094 \times 10^{3} = 2.188 \times 10^{6} \times \frac{z}{4} \Rightarrow z = 2 \Rightarrow He^{+}\text{ ion}$

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