Angular momentum of an electron in the second shell of a $Li^{+2}$ ion is:

If the shortest wavelength of the hydrogen atom in the Lyman series is $x$, then the longest wavelength in the Balmer series of $He^+$ is:

Supposing the electron is present in the $4^{th}$ energy level of $H$ atom. When the electron returns to the ground state,the possible transitions would be:

$A$: Thomson's atomic model is known as the plum pudding model.
$R$: The atom is considered as a positively charged sphere in which electrons are embedded.

The kinetic energy of an electron in the ground state of a $Be^{3+}$ ion is given by:

For the Balmer series in the spectrum of $H$ atom,$\bar{v}=R_{H}\left\{\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right\}$,the correct statements among $(I)$ to $(IV)$ are:
$(I)$ As wavelength decreases,the lines in the series converge.
$(II)$ The integer $n_{1}$ is equal to $2$.
$(III)$ The lines of longest wavelength corresponds to $n_{2}=3$.
$(IV)$ The ionization energy of hydrogen can be calculated from wave number of these lines.