The de-Broglie wavelength of an electron in the ground state of the hydrogen atom is

The speed of the electron in a hydrogen atom in the $n=3$ level is (Planck constant $= 6.6 \times 10^{-34} \ J \ s$):

In an atom,two electrons move around the nucleus in circular orbits of radii $R$ and $4R$. The ratio of the time taken by them to complete one revolution 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.

The difference between the $n^{th}$ and $(n+1)^{th}$ Bohr radius of an $H$ atom is equal to its $(n-1)^{th}$ Bohr radius. The value of $n$ is:

Magnetic field at the centre of the hydrogen atom due to the motion of an electron in the $n^{\text{th}}$ orbit is proportional to: