For the Balmer series, in the spectrum of $\mathrm{H}$ atom, $\stackrel{-}{\mathrm{v}}={\mathrm{R}}_{\mathrm{H}}\left\{\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right\},$ the correct statements among $\left(\mathrm{I}\right)$ to $\left(\mathrm{I}\mathrm{V}\right)$ are,
$\left(\mathrm{I}\right)$ As wavelength decreases, the lines in the series converge.
$\left(\mathrm{I}\mathrm{I}\right)$ The integer ${\mathrm{n}}_1$ is equal to $2$.
$\left(\mathrm{I}\mathrm{I}\mathrm{I}\right)$ The lines of the longest wavelength correspond to ${\mathrm{n}}_2=3$.
$\left(\mathrm{I}\mathrm{V}\right)$ The ionization energy of hydrogen can be calculated from the wave number of these lines.

If $\mathrm{p}$ is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength $\mathrm{\lambda },$ then for $1.5\mathrm{ }\mathrm{p}$ momentum of the photoelectron, the wavelength of the light should be:

(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)

The difference between the radii of $3^{\mathrm{rd}}$ and $4^{\text{th }}$ orbits of ${\mathrm{Li}}_{}^{2+}$ is ${\mathrm{\Delta R}}_1$. The difference between the radii of $3^{\mathrm{rd}}$ and $4^{\text{th }}$ orbits of ${\mathrm{He}}^{+}$ is $\Delta {\mathrm{R}}_2.$ Ratio of  $\Delta {\mathrm{R}}_1:\Delta {\mathrm{R}}_2$ is :