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

$A$ sample of hydrogen atoms is in an excited state (all the atoms). The photons emitted from this sample are made to pass through a filter through which light having a wavelength greater than $800 \ nm$ can only pass. Only one type of photon is found to pass through the filter. The sample's initial excited state is: [Take $hc = 1240 \ eV \cdot nm$,ground state energy of hydrogen atom = $-13.6 \ eV$.]

The following diagram indicates the energy levels of a certain atom. When the system moves from the $2E$ level to the $E$ level,a photon of wavelength $\lambda$ is emitted. The wavelength of the photon produced during its transition from the $\frac{4E}{3}$ level to the $E$ level is

In the $n^{th}$ orbit,the energy of an electron is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$ for a hydrogen atom. The energy required to excite the electron from the first orbit $(n=1)$ to the second orbit $(n=2)$ will be..... eV.

The ground state energy of the hydrogen atom is $-13.6 \ eV$. The kinetic and potential energy of the electron in the second excited state are,respectively:

Energy of an electron in the second orbit of a hydrogen atom is $E$. The energy of an electron $E_3$ in the third orbit of a helium $(He^+)$ ion will be: