If the ionization potential of an atom is $122.4 \, V$,then the excitation potentials for the first and second excited states are respectively:

The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = - \frac{13.6}{n^2} \ eV$. How much energy in $eV$ is required to move an electron from the first orbit to the third orbit?

An electron collides with a fixed hydrogen atom in its ground state. The hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently,the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The minimum kinetic energy $(K.E.)$ of the colliding electron will be.....$eV$.

The energy required to remove an electron in a hydrogen atom from $n = 10$ state is (in $\text{ eV}$)

The energy required to excite an electron from the ground state of a hydrogen atom to the first excited state is:

$A$ singly ionised helium atom in an excited state $(n=4)$ emits a photon of energy $2.6 \,eV$. Given that the ground state energy of hydrogen atom is $-13.6 \,eV$,the energy $E_f$ and quantum number $n$ of the resulting state are respectively,