$A$ difference of $2.3\; eV$ 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?

The energy of a hydrogen atom in the $n^{th}$ orbit is $E_n$. What will be the energy in the $n^{th}$ orbit of a singly ionized helium atom?

The ratio of energies of photons produced due to transition of an electron in a hydrogen atom from the second energy level to the first energy level and from the fifth energy level to the second energy level is:

An electron of a hydrogen-like atom,having $Z=4$,jumps from the $4^{\text{th}}$ energy state to the $2^{\text{nd}}$ energy state. The energy released in this process will be $......... \text{eV}$.
(Given $Rch = 13.6 \text{ eV}$)
Where $R =$ Rydberg constant,
$c =$ Speed of light in vacuum,
$h =$ Planck's constant.

An electron jumps from the $4th$ orbit to the $2nd$ orbit of a hydrogen atom. Given the Rydberg constant $R = 10^5 \text{ cm}^{-1}$. The frequency in $\text{Hz}$ of the emitted radiation will be:

In the third orbit of a hydrogen atom,the energy of an electron is $E$. In the fifth orbit of a helium ion $(Z=2)$,the energy of an electron will be: