If an electron jumps from the $1^{st}$ orbital to the $3^{rd}$ orbital,then it will:

The figure below is the plot of potential energy versus internuclear distance $(d)$ of $H_2$ molecule in the electronic ground state. What is the value of the net potential energy $E_0$ (as indicated in the figure) in $kJ \ mol^{-1}$, for $d=d_0$ at which the electron-electron repulsion and the nucleus-nucleus repulsion energies are absent? As reference, the potential energy of $H$ atom is taken as zero when its electron and the nucleus are infinitely far apart.
Use Avogadro constant as $6.023 \times 10^{23} \ mol^{-1}$.

The magnetic moment $(\mu)$ of an electron revolving around the nucleus varies with the principal quantum number $n$ as

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: $[\mu_0 \rightarrow \text{permeability of free space, } m \rightarrow \text{mass of electron, } \varepsilon_0 \rightarrow \text{permittivity of free space, } h \rightarrow \text{Planck's constant}]$

The frequency of the light emitted when an electron transitions from the $n=4$ to $n=2$ level in a hydrogen atom is $\frac{3}{7}$ times the frequency of a transition in a $Li^{2+}$ ion. Which transition in the $Li^{2+}$ ion corresponds to this?

When an electron jumps from the orbit $n=2$ to $n=4$,the wavelength of the radiation absorbed will be ($R$ is Rydberg's constant).