An electron in the hydrogen atom initially in the fourth excited state makes a transition to $n^{\text{th}}$ energy state by emitting a photon of energy $2.86 \ eV$. The integer value of $n$ will be . . . . . . .

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 the ionization energy of a Bohr's hydrogen atom to that of a Bohr's hydrogen-like lithium atom is:

When an electron falls from a higher energy level to a lower energy level,the difference in energy is emitted as electromagnetic radiation. Why can it not be emitted as other forms of energy?

The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is

Assuming the atom is in the ground state,the expression for the magnetic field at the nucleus in a hydrogen atom due to the circular motion of the electron is: $[\mu_0 \rightarrow \text{permeability of free space, } m \rightarrow \text{mass of electron, } \varepsilon_0 \rightarrow \text{permittivity of free space, } h \rightarrow \text{Planck's constant}]$