In a hydrogen-like atom with atomic number $Z$,an electron is in an excited state with principal quantum number $2n$. The maximum energy of a photon that can be emitted from this state is $204 \ eV$. If the electron transitions from the $2n$ orbit to the $n$ orbit,a photon with energy $40.8 \ eV$ is emitted. The value of $n$ is:

In $Li^{++}$ ion,an electron in the first Bohr orbit is excited to a higher energy level by a radiation of wavelength $\lambda$. When the ion de-excites to the ground state in all possible ways (including intermediate emissions),a total of six spectral lines are observed. What is the value of $\lambda$ in $nm$? (Given: $h = 6.63 \times 10^{-34} \, J \cdot s, c = 3 \times 10^8 \, m/s, 1 \, eV = 1.6 \times 10^{-19} \, J$)

$A$ monochromatic radiation of wavelength $\lambda$ is incident on a hydrogen sample in ground state. The sample subsequently emits radiation of six different wavelengths, then the value of $\lambda$ is. [Use $hc = 1242 \text{ eV-nm}$] (in $\text{ nm}$)

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:

Explain Bohr's atomic model.

The angular momentum of an electron in a hydrogen atom is proportional to: (Where $r$ is the radius of the orbit of the electron)