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  4. Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its ground state, on the basis of Bohr’s model, will be about

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Q. Taking the Bohr radius as $a_0 = 53\, pm$, the radius of $Li^{++}$ ion in its ground state, on the basis of Bohr’s model, will be about

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Solution:

Here, $a_0 = 53$ pm, $n = 1$ for ground state
For $Li^{++} $ ion, $Z = 3$
Radius of $n^{th}$ orbit
$ r = \frac{n^2 h^2}{4\pi^2 \,m KZe^2} = \frac {a_0 n^2}{Z}$
$\therefore r = \frac{53\times (1)^2}{3} $
$ [\because a_0 = \frac {h^2}{4\pi^2\,mKe^2} = 53\,pm]$
$ = 17.66 \approx 18 \,pm$