Calculating the ionization energy for hydrogen What energy would be needed to remove the electron from the $n = 4$ level of the hydrogen atom?
  
  
  $\pu{−3.49 * 10^{−17} J}$
  $\pu{−1.36 * 10^{−19} J}$
  $\pu{+2.18 * 10^{−18} J}$
  $\pu{+1.36 * 10^{−19} J}$
  


I assume that the way to do this is to start from the Rydberg formula,

$$
E = \mathcal{R}Z^2 \left( \frac{1}{n_i^2}-\frac{1}{n_f^2} \right),
$$

set the initial level of the electron as $n_i = 4$, and the final level corresponding to removing the electron (ionization) as $n_f = \infty \implies \frac{1}{\infty^2} = 0$, leading to

$$
E = \pu{13.6*\frac{1}{4^2} eV = 0.85 eV = 1.36*10^{-19} J}.
$$
I assume you refer to Bohr's atomic model.

The energy of an electron in Bohr’s orbit of Hydrogen atom is given by the expression:

$$
\begin{align}
E_{n} &amp;= \frac{2\pi^{2}me^{4}Z^{2}}{n^{2}h^{2}\left(4\pi\epsilon_{0}\right)^{2}} \\
&amp;= \color{\navy}{-13.6\frac{Z^2}{n^2}\pu{ eV}}
\end{align}
$$

Since $Z = 1$ for hydrogen, the above equation can be further simplified. Now just plug in the required values.