Question

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

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

Answer 1

The correct option is B 91.8 eV

For $L{i}^{++}$ model,

$2=3$

In first excitation of ground state,

$n_1=1$ and $n_2=2$

In Bohr's model,

$\Delta E=\left(13.6ev\right)2^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

Here, $\Delta E=\left(13.6\right)(3{)}^2[\frac{1}{1}-\frac{1}{4}]$

$\Delta E=\left(13.6\right)9\times \frac{3}{4}=91.8$eV

So, first excitation energy for ground state in $L{i}^{++}$ is $91.8$eV

Option B is correct.