If the speed of an electron of a hydrogen atom in the ground state is $2.2 \times 10^6 \ m/s$,then its speed in the third excited state will be

$A$ sample of hydrogen-like atoms produces an emission spectrum consisting of $10$ wavelengths arising from all possible transitions. During this process,the maximum angular momentum change for an electron transitioning from a higher energy level to a lower energy level is:

The orbital frequency of an electron in the hydrogen atom is proportional to

$A$ 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 $.80$)

Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level $n$ to level $(n-1)$. For large $n$,show that this frequency equals the classical frequency of revolution of the electron in the orbit.

In Bohr's atomic model,the electron is assumed to revolve in a circular orbit of radius $0.5 \times 10^{-10} \; m$. If the speed of the electron is $2.2 \times 10^{6} \; m/s$,then the current associated with the electron will be $.... \times 10^{-2} \; mA$. [Take $\pi$ as $\frac{22}{7}$]