The diagram below shows the lowest four energy levels for an electron in a hypothetical atom. The electron is excited to the $-1\,eV$ level and transitions to the lowest energy state $(-12\,eV)$ by emitting exactly two photons. Which of the following energies could not belong to either of the photons (in $eV$)?

$A$ 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,

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,

$A$ hydrogen atom is bombarded with electrons accelerated through a potential difference of $V$, which causes excitation of hydrogen atoms. If the experiment is performed at $T = 0 \,K$, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $\frac{\alpha}{10} \,V$, where $\alpha = $ . . . . . . .

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with maximum wavelength?

The following diagram indicates the energy levels of a certain atom. When the system moves from the $4E$ level to the $E$ level,a photon of wavelength ${\lambda _1}$ is emitted. The wavelength of the photon produced during its transition from the $\frac{7}{3}E$ level to the $E$ level is ${\lambda _2}$. The ratio $\frac{{{\lambda _1}}}{{{\lambda _2}}}$ will be: