Question: Taking the Bohr radius as $q_{0} = 53$ the radius of $Li^{+ +}$ion in its ground state, on the b...

Question

Taking the Bohr radius as $q_{0} = 53$ the radius of $Li^{+ +}$ ion in its ground state, on the basis of Bohr’s model, will be about

Answer 1

Here, $a_{0} = 53$ pm, $n = 1$ for ground state

For $Li^{+ +}ion,Z = 3$

Radius of nth orbit

$r = \frac{n^{2}h^{2}}{4\pi^{2}mKZe^{2}} = \frac{a_{0}n^{2}}{Z}$

$\therefore r = \frac{53 \times (1)^{2}}{3}\left\lbrack \therefore a_{0} = \frac{h^{2}}{4\pi^{2}mKe^{2}} = 53pm \right\rbrack$

$= 17.66 \approx 18pm$

Question: Taking the Bohr radius as \(q_{0} = 53\) the radius of \(Li^{+ +}\)ion in its ground state, on the b...