What is the radius of an iodine atom (atomic number $53$,mass number $126$)?

According to Bohr's theory,the angular momentum of an electron revolving in the second orbit of a hydrogen atom will be:

The inverse square law in electrostatics is $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}{r^2}}}$ for the force between an electron and a proton. The $\frac{1}{r^2}$ dependence of $|\vec F|$ can be understood in quantum theory as being due to the fact that the particle of light (photon) is massless. If photons had a mass $m_p$,the force would be modified to $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}}}\left( {\frac{1}{{{r^2}}} + \frac{\lambda }{r}} \right)\left( {{e^{ - \lambda r}}} \right)$ where $\lambda = \frac{{{m_p}c}}{\hbar }$ and $\hbar = \frac{h}{{2\pi }}$. Estimate the change in the ground state energy of a $H$-atom if $m_p$ were $10^{-6}$ times the mass of an electron.

The de-Broglie wavelength associated with an electron of a hydrogen atom in its ground state is (in $\text{ Å}$)

$A$ hydrogen atom in its ground state absorbs $10.2 \ eV$ of energy. The orbital angular momentum is increased by (Given Planck constant $h = 6.6 \times 10^{-34} \ J \cdot s$)

Two electrons are moving in orbits of two hydrogen-like atoms with speeds $3 \times 10^5 \ m/s$ and $2.5 \times 10^5 \ m/s$ respectively. If the radii of these orbits are nearly the same, then the possible order of energy states are . . . . . . respectively.