The de Broglie wavelength of the electron in the ground state is $\lambda_1$ and that in the $n = 3$ level is $\lambda_3$. Then $\lambda_3$ is given by:

In a hydrogen atom,if an electron in the orbit with principal quantum number $n$ jumps to the first excited state,the wavelength of the emitted photon is $\lambda$. Then the value of $n$ is (where $R$ is the Rydberg constant).

An electron in the hydrogen atom excites from the $2^{nd}$ orbit to the $4^{th}$ orbit. The change in angular momentum of the electron is (Planck's constant $h = 6.64 \times 10^{-34} \ J \cdot s$).

When an electron in a hydrogen atom jumps from the third excited state to the ground state,the de-Broglie wavelength associated with the electron becomes

$A$ particle of mass $m$ moves around the origin in a potential $\frac{1}{2} m \omega^{2} r^{2}$,where $r$ is the distance from the origin. Applying the Bohr's model in this case,the radius of the particle in its $n$th orbit in terms of $a=\sqrt{\frac{h}{2 \pi m \omega}}$ is

Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?