If the first excitation potential of a hydrogen-like atom is $V \, \text{volt}$, then the ionization energy of this atom will be

The angular momentum of an electron in a hydrogen atom is proportional to: (Where $r$ is the radius of the orbit of the electron)

If the radius of the first Bohr orbit is $r$,then the de-Broglie wavelength of the electron in the $4^{\text{th}}$ orbit will be:

For $He^{+}$, a transition takes place from the orbit of radius $105.8 \ pm$ to the orbit of radius $26.45 \ pm$. The wavelength (in $nm$) of the emitted photon during the transition is. . . . .
[Use: Bohr radius, $a_0=52.9 \ pm$; Rydberg constant, $R_H=2.2 \times 10^{-18} \ J$; Planck's constant, $h=6.6 \times 10^{-34} \ J \ s$; Speed of light, $c=3 \times 10^8 \ m \ s^{-1}$]

An electron in a hydrogen atom,after absorbing energy photons,can jump between energy states $n_1$ and $n_2$ $(n_2 > n_1)$. It then returns to the ground state,emitting six different wavelengths in the emission spectrum. The energy of the emitted photons can be equal to,less than,or greater than the absorbed photon energy. Determine $n_1$ and $n_2$.

An orbital electron in the ground state of hydrogen has a magnetic moment $\mu_1$. This orbital electron is excited to the $3^{rd}$ excited state by some energy transfer to the hydrogen atom. If the new magnetic moment of the electron is $\mu_2$,then: