For ${\mathrm{He}}^{+}$, a transition takes place from the orbit of radius $105.8\mathrm{pm}$ to the orbit of radius $26.45\mathrm{pm}$. The wavelength (in $\mathrm{nm}$ ) of the emitted photon during the transition is

Bohr radius, $\mathrm{a}=52.9 \mathrm{pm}$
Rydberg constant, ${\mathrm{R}}_{\mathrm{H}}=2.2\times {10}^{-18} \mathrm{J}$
Planck's constant, $\mathrm{h}=6.6\times {10}^{-34}\mathrm{Js}$
Speed of light, $\mathrm{c}=3\times {10}^8 {\mathrm{ms}}^{-1}$ ]

The figure below is the plot of potential energy versus internuclear distance $\left(\mathrm{d}\right)$ of ${\mathrm{H}}_2$ molecule in the electronic ground state. The value of the net potential energy ${\mathrm{E}}_0$ (as indicated in the figure) is $-5.24649\times {10}^{\mathrm{n}}$ $\mathrm{kJ} {\mathrm{mol}}^{-1},$ for $\mathrm{d}={\mathrm{d}}_0$ at which the electron-electron repulsion and the nucleus-nucleus repulsion energies are absent, find the value of $\mathrm{n}$ to the nearest integer value.

As reference, the potential energy of $\mathrm{H}$ atom is taken as zero when its electron and the nucleus are infinitely far apart.

Use Avogadro constant as $6.023\times {10}^{23} {\mathrm{mol}}^{-1}.$  Give an answer to the nearest integer value.

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 :