Question

Taking the Bohr radius as $a_0=35pm$, the radius of $L{i}^{++}$ion in its ground state, on the basis of Bohr's model, will be about

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

Answer 1

The correct option is C $18pm$

$\begin{array}{c}\text{Given,}\\ \text{Bohr's radius}=53pm\\ \text{we know,}\\ \text{Bohr's radius}=\frac{0.529n^2}{2}\end{array}$

$\begin{array}{cc}\text{if}n=1& Z=1\text{, the radius obtained is bohr's radius}\end{array}$

$\begin{array}{c}\text{radius}\left(a_0\right)=\frac{0.529(1{)}^2}{1}A=53Pm\\ \text{for}L{i}^{++}n=1\left(\text{ground state}\right)\\ \text{and}z=3\end{array}$

$\begin{array}{cc}\text{radius}& =\frac{53\cdot n^2}{2}{p}_m=\frac{53(1{)}^2}{3}\\ r& =17.6\approx 18pm\\ \text{hence option (c) is wrrect.}& \end{array}$