In a Bohr atom,the total energy of an electron in the $n$-th allowed orbit is $E_n$ and its angular momentum is $J_n$. Then:

The potential energy of an electron in an orbit of a hydrogen atom is $-6.8 \text{ eV}$. The de Broglie wavelength of the electron in this orbit is (where $r_0$ is the Bohr radius).

The orbital frequency of an electron in the hydrogen atom is proportional to

$A$ hydrogen atom is in the ground state. In order to get six lines in its emission spectrum,the wavelength of the incident radiation should be .......... $\mathring A$.

Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr model to this atom. The longest wavelength photon that will be emitted has wavelength $\lambda$ (given in terms of the Rydberg constant $R$ for the hydrogen atom) equal to:

$A$ hydrogen atom in its ground state absorbs $10.2 \ eV$ of energy. The orbital angular momentum is increased by (Given Planck constant $h = 6.6 \times 10^{-34} \ J \cdot s$)