Question

Assuming Bohr's model for $L{i}^{++}$ atom, the first excitation energy of ground state of $L{i}^{++}$ atom is

• A. $10.2eV$
• B. $91.8eV$
• C. $13.6eV$
• D. $3.4eV$

Answer 1

The correct option is B $91.8eV$

$\begin{array}{c}\text{Given,}\\ L{i}^{++}\text{atom:}\\ \text{first excitation Energy of ground state}L{i}^{++}\\ \text{mean excitation of electrom from}\\ n=1\text{to}n=2\end{array}$

$\begin{array}{c}\text{we know,}\\ \text{Energy}=13.6z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\end{array}$

$\begin{array}{cc}& =13.6\times 9\times (\frac{1}{1}-\frac{1}{4})\\ E& =\frac{13.6.\times 9\times 3}{4}ev\\ E& =91.8ev\end{array}$