An electron jumps from the $5^{th}$ orbit to the $4^{th}$ orbit of a hydrogen atom. Taking the Rydberg constant as $10^7 \ m^{-1}$,what will be the frequency of the radiation emitted?

In a hydrogen atom, the energy of an electron in the first and third orbits is $E_1$ and $E_3$ respectively. If $E_3 = x E_1$, then the value of $x$ will be:

The ratio of the energies of the hydrogen atom in its first to second excited state is

The wavelength of the radiation emitted is $\lambda_0$ when an electron jumps from the second excited state to the first excited state of a hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom,the wavelength of the radiation emitted will be $\frac{20}{x} \lambda_0$. The value of $x$ is $........$

Which of the transitions in a hydrogen atom emits a photon of the lowest frequency? ($n =$ quantum number)

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.