When a hydrogen atom absorbs a photon of wavelength $60 \ nm$,the atom undergoes photo-ionization. What will be the maximum kinetic energy of the emitted electron in $eV$?

In the following atoms and molecules,for the transition from $n = 2$ to $n = 1$,the spectral line of minimum wavelength will be produced by:

Consider a hydrogen-like atom whose energy in the $n^{th}$ excited state is given by $E_n = - \frac{13.6 Z^2}{n^2}$. When this excited atom makes a transition from an excited state to the ground state, the most energetic photons have energy $E_{max} = 52.224 \ eV$ and the least energetic photons have energy $E_{min} = 1.224 \ eV$. The atomic number of the atom is:

In $Li^{++}$ ion,an electron in the first Bohr orbit is excited to a higher energy level by a radiation of wavelength $\lambda$. When the ion de-excites to the ground state in all possible ways (including intermediate emissions),a total of six spectral lines are observed. What is the value of $\lambda$ in $nm$? (Given: $h = 6.63 \times 10^{-34} \, J \cdot s, c = 3 \times 10^8 \, m/s, 1 \, eV = 1.6 \times 10^{-19} \, J$)

Imagine that the electron in a hydrogen atom is replaced by a muon $(\mu)$. The mass of the muon particle is $207$ times that of an electron and its charge is equal to the charge of an electron. The ionization potential of this hydrogen atom will be ............. $eV$.

Magnetic moment due to the motion of the electron in the $n$th energy state of a hydrogen atom is proportional to ..........