When an electron in a hydrogen atom makes a transition from the $2^{nd}$ excited state to the ground state,it emits a photon of frequency $f$. The frequency of the photon emitted when an electron of $Li^{++}$ makes a transition from the $1^{st}$ excited state to the ground state is:

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.

Minimum excitation potential of Bohr's first orbit in hydrogen atom is.....$V$

Assertion : Bohr had to postulate that the electrons in stationary orbits around the nucleus do not radiate.
Reason : According to classical physics all moving electrons radiate.

Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the $2^{nd}$ and $4^{th}$ orbits of hydrogen atom. (in $32$ : $1$)

Positronium is just like a $H$-atom with the proton replaced by the positively charged antiparticle of the electron (called the positron,which is as massive as the electron). What would be the ground state energy of positronium?