The speed of the electron in a hydrogen atom in the $n=3$ level is (Planck constant $= 6.6 \times 10^{-34} \ J \ s$):

$A$ diatomic molecule is made of two masses $m_1$ and $m_2$ which are separated by a distance $r$. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization,its energy will be given by: ($n$ is an integer)

$A$ moving Hydrogen atom makes a head-on collision with a stationary Hydrogen atom. Before the collision,both atoms are in the ground state,and after the collision,they move together. The minimum kinetic energy of the moving Hydrogen atom,such that one of the atoms reaches the excitation state,is (in $eV$)

The ratio of areas within the electron orbits for the first excited state to the ground state for a hydrogen atom is (in $ : 1$)

According to Bohr's theory,the time-averaged magnetic field at the centre (i.e.,nucleus) of a hydrogen atom due to the motion of electrons in the $n^{th}$ orbit is proportional to ($n =$ principal quantum number).

According to de-Broglie, the de-Broglie wavelength for an electron in an orbit of a hydrogen atom is $10^{-9} \, m$. The principal quantum number for this electron is: