An electron makes a transition from orbit $n = 4$ to the orbit $n = 2$ of a hydrogen atom. The wave number of the emitted radiations ($R =$ Rydberg's constant) will be

Mention the value of the Rydberg constant with its unit.

The ionisation potential of $H$-atom is $13.6 \, eV$. When it is excited from the ground state by monochromatic radiation of $970.6 \, \mathring{A}$,the number of emission lines will be (according to Bohr's theory):

The de-Broglie wavelength $(\lambda_B)$ associated with the electron orbiting in the second excited state of a hydrogen atom is related to that in the ground state $(\lambda_G)$ by:

$A$ hydrogen-like atom of atomic number $Z$ is in an excited state of quantum number $2n$. It can emit a maximum energy photon of $204 \ eV$. If it makes a transition to quantum state $n$,a photon of energy $40.8 \ eV$ is emitted. The value of $n$ will be

Imagine removing one electron from $He^4$ and $He^3$. Their energy levels,as worked out on the basis of the Bohr model,will be very close. Explain why?