In any Bohr orbit of a hydrogen atom,the ratio of $K.E.$ to $P.E.$ of a revolving electron at a distance $r$ from the nucleus is:

The wavelength of light emitted when an electron transitions from the second orbit to the first orbit in a hydrogen atom is:

Radius of the first orbit of the electron in a hydrogen atom is $0.53 \; \mathring{A}$. So,the radius of the third orbit will be ........... $\mathring{A}$.

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

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 ground state energy of a hydrogen atom is $-13.6 \text{ eV}$. The potential energy of the electron in this state is: (in $\text{ eV}$)