Energy required for the electron excitation in $Li^{2+}$ from the first to the third Bohr orbit is
Answer & explanation
Correct answer: option 2
$\Delta E = 13.6 z^2 (\frac{1}{n^2} - \frac{1}{m^2})$
$\Delta E = 13.6 3^2 (\frac{1}{1^2} - \frac{1}{3^2}) = -13.6 \times 9 (1 - \frac{1}{9}) = 13.6 \times 8 = 108.8 J$