The energy $(E)$ of the $n^{\text{th}}$ Bohr orbit of an atom is given by:
$E_{n} = \frac{-(2.18 \times 10^{-18} \ J) Z^{2}}{n^{2}}$
Given,ground state energy $= -2.18 \times 10^{-11} \ erg = -2.18 \times 10^{-18} \ J$.
Energy required to shift the electron from $n=1$ to $n=5$ is:
$\Delta E = E_{5} - E_{1} = -2.18 \times 10^{-18} [\frac{1}{5^{2}} - \frac{1}{1^{2}}]$
$\Delta E = -2.18 \times 10^{-18} [\frac{1}{25} - 1] = -2.18 \times 10^{-18} [-\frac{24}{25}]$
$\Delta E = 2.0928 \times 10^{-18} \ J$.
When the electron returns to the ground state,the energy emitted is equal to the energy absorbed,$\Delta E = 2.0928 \times 10^{-18} \ J$.
The wavelength $(\lambda)$ of the emitted light is given by:
$\lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34} \ J \cdot s) (3 \times 10^{8} \ m/s)}{2.0928 \times 10^{-18} \ J}$
$\lambda = 9.498 \times 10^{-8} \ m$.