$A$ $15.0\, eV$ photon collides with and ionizes a hydrogen atom. If the atom was originally in the ground state (ionization potential $= 13.6\, eV$),what is the kinetic energy of the ejected electron? .......... $eV$

Hydrogen $(H)$,Deuterium $(D)$,Helium $(He^+)$,and Lithium $(Li^{2+})$ emit radiation of wavelengths $\lambda_1, \lambda_2, \lambda_3,$ and $\lambda_4$ respectively during the transition from $n = 2$ to $n = 1$. Then:

The radius of the fifth orbit of the $Li^{++}$ ion is $......... \times 10^{-12} \ m$. Take: radius of the hydrogen atom (first Bohr radius) $r_0 = 0.51 \ \mathring{A}$.

Bohr's atom model assumes

Monochromatic radiation of wavelength $\lambda$ is incident on a hydrogen sample containing atoms in the ground state. Hydrogen atoms absorb the light and subsequently emit radiations of ten different wavelengths. The value of $\lambda$ is.....$nm$.

Assuming the atom is in the ground state, the expression for the magnetic field at the nucleus in a hydrogen atom due to the circular motion of the electron is [$\mu_0 =$ permeability of free space, $m =$ mass of electron, $\epsilon_0 =$ permittivity of free space, $h =$ Planck's constant].