The binding energy of an electron in the ground state of a hydrogen atom is $13.6 \ eV$. What is the energy in $eV$ required to remove an electron from the first excited state of a $Li^{++}$ ion?

$A$ hydrogen atom in its ground state is irradiated by light of wavelength $970 \mathring A$. Taking $hc/e = 1.237 \times 10^{-6} \text{ eV m}$ and the ground state energy of the hydrogen atom as $-13.6 \text{ eV}$,the number of lines present in the emission spectrum is

The energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E = \frac{-13.6}{n^2} \; eV$. The energy of a photon emitted when the electron jumps from the $n = 3$ state to the $n = 2$ state of hydrogen is approximately......$eV$.

$A$ $12.5 \ eV$ electron beam is used to bombard gaseous hydrogen at room temperature. It will emit:

The ionization energy of hydrogen is $13.6 \text{ eV}$. The energy of the photon released when an electron jumps from the first excited state $(n=2)$ to the ground state of a hydrogen atom is (in $\text{ eV}$)

The figure shows the energy levels of a certain atom. When the electron de-excites from $3E$ to $E$,an electromagnetic wave of wavelength $\lambda$ is emitted. What is the wavelength of the electromagnetic wave emitted when the electron de-excites from $\frac{5E}{3}$ to $E$?