A beam of monochromatic light is used to excite the electron in Li...

Energy released in the transition will be,

$\DeltaE = 13 . 6 \times {Z}^{2}$\left[\dfrac{1}{{1}^{2}} - \dfrac{1}{{3}^{2}}\right] = 13 . 6 \times {3}^{2} \times \dfrac{8}{9}eV = 108 . 8eV$</p><p>Now energy of photon can be written as,</p><p>$\dfrac{hc}{\lambda} = 108 . 8eV ⇒ \lambda = \dfrac{12400}{108 . 8}\overset{o}{A} \approx 114\overset{o}{A}