The angular momentum of the orbital electron is an integral multiple of:

The ratio of areas within the electron orbits for the first excited state to the ground state for a hydrogen atom is (in $ : 1$)

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 number of de-Broglie wavelengths contained in the second Bohr orbit of a hydrogen atom is

Imagine removing one electron from $He^4$ and $He^3$. Their energy levels,as worked out on the basis of the Bohr model,will be very close. Explain why?

The ionization potential for the second $He$ electron is ...... $eV$.