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 following diagram indicates the energy levels of a certain atom. When the system moves from the $2E$ level to the $E$ level, it emits a photon of wavelength $\lambda$. What is the wavelength of the photon produced during its transition from the $\frac{4E}{3}$ level to the $E$ level?

The diagram below shows the lowest four energy levels for an electron in a hypothetical atom. The electron is excited to the $-1\,eV$ level and transitions to the lowest energy state $(-12\,eV)$ by emitting exactly two photons. Which of the following energies could not belong to either of the photons (in $eV$)?

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.

$A$ hydrogen sample is prepared in a particular excited state $A$ of quantum number $n_A=3$. The ground state energy of the hydrogen atom is $-|E|$. Photons of energy $\frac{|E|}{12}$ are absorbed by the sample,which results in the excitation of some electrons to an excited state $B$ of quantum number $n_B$. The value of $n_B$ is:

For a certain hypothetical one-electron atom,the wavelength (in $\mathring{A}$) for the spectral lines for transition from $n = p$ to $n = 1$ is given by $\lambda = \frac{1500p^2}{p^2 - 1}$ (where $p > 1$). The ionization potential of this element must be ......$V$ (Take $hc = 12420 \text{ eV-} \mathring{A}$).