When an electron in a hydrogen atom jumps from the third orbit to the second orbit,it emits a photon of wavelength $\lambda$. When it jumps from the fourth orbit to the third orbit,the wavelength emitted by the photon will be

Energy of a stationary electron in the hydrogen atom is $E = -\frac{13.6}{n^2} \text{ eV}$. Calculate the energies required to excite the electron in a hydrogen atom to $(a)$ its second excited state and $(b)$ its ionized state,respectively.

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.

The following diagram indicates the energy levels of a certain atom. When the system moves from the $4E$ level to the $E$ level,a photon of wavelength ${\lambda _1}$ is emitted. The wavelength of the photon produced during its transition from the $\frac{7}{3}E$ level to the $E$ level is ${\lambda _2}$. The ratio $\frac{{{\lambda _1}}}{{{\lambda _2}}}$ will be:

An electron in the $n = 1$ orbit of a hydrogen atom is bound by $13.6 \, eV$. If a hydrogen atom is in the $n = 3$ state,how much energy is required to ionize it?

How much energy is needed to release the electron from the second excited state in the hydrogen atom (in $eV$)?