Consider the spectral line resulting from the transition $n = 2 \rightarrow n = 1$ in the atoms and ions given below. The shortest wavelength is produced by:

When an electron jumps from the orbit $n=2$ to $n=4$,the wavelength of the radiation absorbed will be ($R$ is Rydberg's constant).

Assuming the atom in the ground state,the expression for the magnetic field at a point (nucleus) in a hydrogen atom due to the circular motion of the electron is [$\mu_{0} =$ permeability of free space,$\epsilon_{0} =$ permittivity of free space,$m =$ mass of electron,$e =$ electronic charge,$h =$ Planck's constant].

$A$ diatomic molecule is made of two masses $m_1$ and $m_2$ separated by a distance $r$. Applying the Bohr's quantization rule for angular momentum,calculate its rotational kinetic energy. It is given by the formula:

An electron is revolving in $n = 3$ orbit. What will be the magnetic field at the centre of the hydrogen atom? (in $T$)

The radius of the Bohr orbit in the ground state of a hydrogen atom is $0.5 \ \mathring{A}$. The radius of the orbit of the electron in the third excited state of $He^+$ will be ...... $\mathring{A}$.