An electron in the $n = 1$ orbit of a hydrogen atom is bound by $13.6 \, eV$. If a hydrogen atom is in the $n = 3$ state,how much energy is required to ionize it?

What is an emission line?

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the photons emitted when electrons in the $n^{\text{th}}$ orbit of a hydrogen atom fall to the first excited state and ground state respectively,then the value of $n$ is:

An electron with kinetic energy $E$ collides with a hydrogen atom in the ground state. The collision will be elastic

In a hydrogen atom,if the difference in the energy of the electron in $n = 2$ and $n = 3$ orbits is $E,$ the ionization energy of the hydrogen atom is ....... $E$.

$A$ 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