If the potential energy of a hydrogen atom in the first excited state is assumed to be zero,then the total energy of the $n = \infty$ state is,

Energy of the electron in $n^{th}$ orbit of hydrogen atom is given by $E_n = - \frac{13.6}{n^2} \ eV$. The amount of energy needed to transfer an electron from the first orbit to the third orbit 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:

When a hydrogen atom is raised from the ground state to an excited state,

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?

$A$ 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: