The electron energy in hydrogen atom is given by

Solution:

Given

$E_{n}=-\frac{2.18 \times 10^{-18}}{n^{2}} \mathrm{~J}$

Energy required for ionization from n = 2 is given by,

$\Delta E=E_{\infty}-E_{2}$

$=\left[\left(\frac{-2.18 \times 10^{-18}}{(\infty)^{2}}\right)-\left(\frac{-2.18 \times 10^{-18}}{(2)^{2}}\right)\right] J$

$=\left[\frac{2.18 \times 10^{-18}}{4}-0\right] \mathrm{J}$

$=0.545 \times 10^{-18} \mathrm{~J}$

$\Delta \mathrm{E}=5.45 \times 10^{-19} \mathrm{~J}$

$\lambda=\frac{\mathrm{hc}}{\Delta E}$

Here, $\lambda$ is the longest wavelength causing the transition.

$\lambda=\frac{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{5.45 \times 10^{-19}}=3.647 \times 10^{-7} \mathrm{~m}$

$=3647 \times 10^{-10} \mathrm{~m}$

$=3647 \AA$