In the Bohr model of the hydrogen atom,the linear momentum of the electron depends on the principal quantum number $n$ as:

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 in an excited state transitions to the ground state by emitting a photon of wavelength $\lambda$. The value of the principal quantum number $n$ of the excited state is:
($R$: Rydberg constant)

What is the angular momentum of an electron in Bohr's hydrogen atom whose energy is $-0.544 \ eV$?

$A$ particular hydrogen-like ion emits radiation of frequency $2.92 \times 10^{15} \text{ Hz}$ when it makes a transition from $n=3$ to $n=1$. The frequency in $\text{Hz}$ of radiation emitted in the transition from $n=2$ to $n=1$ will be: (in $\times 10^{15}$)

In the Auger process,an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron). Assuming the nucleus to be massive,calculate the kinetic energy of an $n = 4$ Auger electron emitted by Chromium by absorbing the energy from an $n = 2$ to $n = 1$ transition.