The radius of the orbit of an electron in a hydrogen atom is $0.5 \ \mathring{A}$. The speed of the electron is $2 \times 10^6 \ m/s$. The current in the loop due to the motion of the electron is ............. $mA$.

In the Bohr's hydrogen atom model,the radius of the stationary orbit is directly proportional to ($n =$ principle quantum number)

In the hydrogen atom, the electron makes a transition from the higher orbit $(i)$ to a lower orbit $(f)$. The ratio of the radius of the orbits is given by $r_i : r_f = 16 : 4$. The wavelength of the photon emitted due to this transition is . . . . . . nm. (Given Rydberg constant $R = 1.0973 \times 10^7 \text{ m}^{-1}$)

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

If $m$ is the mass of an electron,$v$ is its velocity,$r$ is the radius of a stationary circular orbit around a nucleus with charge $Ze$,then from Bohr's first postulate,the kinetic energy of the electron is (where $K = 1 / 4 \pi \epsilon_0$):

The magnetic moment of an electron due to its orbital motion is proportional to ($n$ = principal quantum number).