The orbital acceleration of an electron is .........

Consider a hydrogen-like ionized atom with atomic number $Z$ with a single electron. In the emission spectrum of this atom,the photon emitted in the $n = 2$ to $n = 1$ transition has energy $74.8 \ eV$ higher than the photon emitted in the $n = 3$ to $n = 2$ transition. The ionization energy of the hydrogen atom is $13.6 \ eV$. The value of $Z$ 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.

When an electron in a Hydrogen atom revolves in a stationary orbit,it

In a hydrogen atom in its ground state,the first Bohr orbit has radius $r_1$. The electron's orbital speed becomes one-third when the atom is raised to one of its excited states. The radius of the orbit in that excited state is (in $r_1$)

In the first excited state of a hydrogen atom, the energy of its electron is $-3.4 \text{ eV}$. The radial distance of the electron from the hydrogen nucleus in this case is approximately: (Take $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$, $e = 1.6 \times 10^{-19} \text{ C}$ and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$)