According to Bohr's model, the highest kinetic energy is associated with the electron in the

Using the formula for the radius of the $n^{th}$ orbit $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$,derive an expression for the total energy of an electron in the $n^{th}$ Bohr orbit.

The ratio of the ionization energy of a Bohr's hydrogen atom to that of a Bohr's hydrogen-like lithium atom is:

An electron in a hydrogen atom,after absorbing energy photons,can jump between energy states $n_1$ and $n_2$ $(n_2 > n_1)$. It then returns to the ground state,emitting six different wavelengths in the emission spectrum. The energy of the emitted photons can be equal to,less than,or greater than the absorbed photon energy. Determine $n_1$ and $n_2$.

An electron moves in a Bohr orbit. The magnetic field at the centre is proportional to

A hydrogen atom, initially in the ground state, is excited by absorbing a photon of wavelength $980 \ \text{\AA}$. The radius of the atom in the excited state, in terms of Bohr radius $a_0$, will be (given $hc = 12500 \ \text{eV-\AA}$). (in $a_0$)