The electron energy in a hydrogen atom is given by $E_n = (-2.18 \times 10^{-18})/n^2 \ J$. Calculate the energy required to remove an electron completely from the $n = 2$ orbit. What is the longest wavelength of light in $cm$ that can be used to cause this transition?

(A) Given,$E_n = -\frac{2.18 \times 10^{-18}}{n^2} \ J$.
Energy required for ionization from $n = 2$ is $\Delta E = E_{\infty} - E_2$.
Since $E_{\infty} = 0$,$\Delta E = 0 - (\frac{-2.18 \times 10^{-18}}{2^2}) = \frac{2.18 \times 10^{-18}}{4} = 5.45 \times 10^{-19} \ J$.
Using $\lambda = \frac{hc}{\Delta E}$,where $h = 6.626 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$:
$\lambda = \frac{(6.626 \times 10^{-34} \ J \cdot s)(3 \times 10^8 \ m/s)}{5.45 \times 10^{-19} \ J} = 3.647 \times 10^{-7} \ m$.
Converting to $cm$: $\lambda = 3.647 \times 10^{-7} \ m \times 100 \ cm/m = 3.647 \times 10^{-5} \ cm$.