$A$ beam of monochromatic light is used to excite the electron in $Li^{++}$ from the first orbit to the third orbit. The wavelength of monochromatic light is found to be $x \times 10^{-10} \; m$. The value of $x$ is $\dots$. [Given $hc = 1242 \; eV \cdot nm$]

The angular momentum of the orbital electron is an integral multiple of:

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

An electron in the ground state of a hydrogen atom is revolving in a circular orbit of radius $R$. The orbital magnetic moment of the electron is ($m =$ mass of electron,$h =$ Planck's constant,$e =$ electronic charge).

$A$ hydrogen atom,initially at rest in its ground state,absorbs a photon of frequency $v_1$ and ejects the electron with a kinetic energy of $10 eV$. The electron then combines with a positron at rest to form a positronium atom in its ground state and simultaneously emits a photon of frequency $v_2$. The center of mass of the resulting positronium atom moves with a kinetic energy of $5 eV$. It is given that the positron has the same mass as that of the electron and the positronium atom can be considered as a Bohr atom,in which the electron and the positron orbit around their center of mass. Considering no other energy loss during the whole process,the difference between the two photon energies (in $eV$) is $....$ (in $.80$)

$A$ 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