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  4. The radius of Li++ ion in its ground state in terms of Bohr radius a0 is

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Q. The radius of $Li^{++}$ ion in its ground state in terms of Bohr radius $a_0$ is

Atoms

Solution:

Radius of $n^{th}$ orbit of hydrogen-like atom is
$r_{n} = \frac{n^{2}}{Z} a_{0}$
For $Li^{++}$ ion, $Z = 3,\, n = 1$ for ground state
$\therefore $ Radius of $Li^{++}$ ion in its ground state is
$r_{1} = \frac{\left(1\right)^{2}}{3}a_{0} = \frac{a_{0}}{3}$