The longest wavelength of light that can be used for the ionisation...

Electronic configuration of $Li = 1{s}^{2}2{s}^{1}$

${\left({E}_{Li}\right)}_{n = 2} = \left({E}_{H}\right)\dfrac{{Z}^{2}}{{n}^{2}} = - 2 . 2 \times {10}^{- 18} \times \dfrac{9}{4}$

$\dfrac{6 . 63 \times {10}^{- 34} \times 3 \times {10}^{8}}{\lambda} = 2 . 2 \times {10}^{- 18} \times \dfrac{9}{4}$

$\lambda = 4 \times {10}^{- 8}m$