In a hydrogen atom,the electron is in the $n^{th}$ excited state. It may come down to the second excited state by emitting ten different wavelengths. What is the value of $n$?

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number $n_i$ to another with quantum number $n_f$. $V_i$ and $V_f$ are respectively the initial and final potential energies of the electron. If $\frac{V_i}{V_f} = 6.25$,then the smallest possible $n_f$ is:

$A$ hydrogen atom (ionisation potential $13.6 \ eV$) makes a transition from the third excited state to the first excited state. The energy of the photon emitted in the process is.......$eV$.

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:

The ionisation potential of a hydrogen atom is $13.6 \, V$. The energy required to remove an electron in the $n = 2$ state of the hydrogen atom is.....$eV$.

During the transition of an electron from state $A$ to state $C$ of a Bohr atom,the wavelength of the emitted radiation is $2000 \ \mathring A$ and it becomes $6000 \ \mathring A$ when the electron jumps from state $B$ to state $C.$ Then the wavelength of the radiation emitted during the transition of electrons from state $A$ to state $B$ is