Question

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

• A. $53$ pm
• B. $27$ pm
• C. $18$ pm
• D. $13$ pm

Answer 1

Answer: (C)

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