The angular momentum for the electron in the first Bohr orbit is $L$. If the electron is assumed to revolve in the second orbit of a hydrogen atom,then the change in angular momentum will be:

The ratio of angular momentum of an electron in $n^{\text{th}}$ orbit of a hydrogen atom to the velocity of the electron in the $n^{\text{th}}$ orbit is proportional to

The frequency of the $1^{st}$ line of the Balmer series in the ${H_2}$ atom is ${\nu _0}$. What is the frequency of the corresponding line emitted by a singly ionized ${He^+}$ atom?

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).

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 a hydrogen atom and a $Li^{2+}$ ion,the electron is in the second excited state. If $l_{H}$ and $l_{Li}$ are the angular momenta of the electrons and $E_H$ and $E_{Li}$ are their respective energies,then: