The ionization energy of the hydrogen atom is $13.6 \ eV$. The potential energy of the electron in $n=2$ state of the hydrogen atom is

The energy of a hydrogen atom in its ground state is $-13.6 \,eV$. The energy of the level corresponding to $n=7$ is ...... $eV$.

An electron in a hydrogen atom first jumps from the third excited state to the second excited state and then from the second excited state to the first excited state. The ratio of the wavelengths $\lambda_1 : \lambda_2$ emitted in the two cases is

If the binding energy of the electron in a hydrogen atom is $13.6 \ eV$,the energy required to remove the electron from the first excited state of $Li^{++}$ is.....$eV$.

If the ionization energy of a hydrogen atom in the ground state is $13.6 \ eV$,find the ionization potential of the third orbit in $eV$.

Energy of an electron in the second orbit of a hydrogen atom is $E_{2}$. The energy of an electron in the third orbit of $He^{+}$ will be