The shortest wavelength in the hydrogen spectrum is approximately (in $nm$)

Four lowest energy levels of $H$-atom are shown in the figure. The number of emission lines could be obtained by transitions from these levels. The total number of possible emission lines is:

$A$ photon of wavelength $4 \times 10^{-7} \, m$ strikes on a metal surface. The work function of the metal is $2.13 \, eV$. Calculate:
$(i)$ The energy of the photon in $eV$.
$(ii)$ The kinetic energy of the emission in $eV$.
$(iii)$ The velocity of the photoelectron in $ms^{-1}$ ($1 \, eV = 1.6020 \times 10^{-19} \, J$,mass of electron $m = 9.10939 \times 10^{-31} \, kg$).

For a $Bohr$ atom,the angular momentum of an electron is given by $M(n = 1, 2, 3, .....)$ :

An electronic transition in a hydrogen atom results in the formation of the $H_\alpha$ line of the hydrogen spectrum in the Lyman series. The energies associated with the electron in each of the orbits involved in the transition (in $kcal \ mol^{-1}$) are:

The ratio of the velocity of an electron in the ground state of a hydrogen atom to its velocity in the second excited state of a $He^{+}$ ion is: