An excited hydrogen atom emits a photon of wavelength $\lambda$ in returning to the ground state. The quantum number $n$ of the excited state is ($R=$ Rydberg's constant).

According to Bohr's atomic model,which of the following is $NOT$ a possible energy for a photon emitted by a hydrogen atom (in $;eV$)? (in $eV$)

If $E$ and $L$ denote the magnitudes of total energy and of angular momentum of an electron in a Bohr orbit,then the true relation between them is:

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).

$A$ hydrogen atom,a deuteron atom,a $He^+$ ion,and a $Li^{++}$ ion each contain a single electron orbiting their nucleus. The wavelengths of the electromagnetic radiation emitted during the transition of the electron from the $n = 2$ orbit to the $n = 1$ orbit are found to be $\lambda_1, \lambda_2, \lambda_3$,and $\lambda_4$ respectively. Then:

The de-Broglie wavelength of an electron in the $3^{rd}$ orbit of a $He^{+1}$ ion is approximately (in $\mathring{A}$):