The ratio of the centripetal accelerations of the electron in two successive orbits of hydrogen is $81: 16$. Due to a transition between these two states,the angular momentum of the electron changes by ($h$ - Planck's constant).

According to Bohr's theory,the angular momentum of an electron revolving in the second orbit of a hydrogen atom will be:

What will be the angular momentum of an electron,if the energy of this electron in an $H$-atom is $-1.5 \, eV$ (in $J \cdot s$)?

In a hydrogen-like ion,the energy difference between the $2^{\text{nd}}$ excitation state and the ground state is $108.8 \ eV$. The atomic number of the ion is:

The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is $...... \text{ } \mathring{A}$ (radius of the first orbit of hydrogen atom $= 0.53 \text{ } \mathring{A}$).

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: $[\mu_0 \rightarrow \text{permeability of free space, } m \rightarrow \text{mass of electron, } \varepsilon_0 \rightarrow \text{permittivity of free space, } h \rightarrow \text{Planck's constant}]$