The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \, eV$. How much energy in $eV$ is required to move an electron from the $n = 2$ orbit to the $n = 3$ orbit?

$A$ difference of $5.4 \ eV$ separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level? (Take $1 \ eV = 1.6 \times 10^{-19} \ J$,$h = 6.625 \times 10^{-34} \ Js$)

Energy of an electron in $n^{th}$ orbit of hydrogen atom is $\left( {k = \frac{1}{{4\pi {\varepsilon _0}}}} \right)$

The energy levels of a hydrogen atom are shown below. The transition corresponding to the emission of the shortest wavelength is

If the electron in a hydrogen atom moves from the ground state orbit to the $5^{\text{th}}$ orbit,then the potential energy of the electron

If $13.6 \ eV$ energy is required to ionize the hydrogen atom,then the energy required to remove an electron from $n=2$ is (in $eV$)