Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the $2^{nd}$ and $4^{th}$ orbits of hydrogen atom. (in $32$ : $1$)

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 frequency of light emitted,when the electron makes a transition from the level of principal quantum number $n=2$ to the level with $n=1$ is (Take,the ionization energy of hydrogen to be $13.6 \ eV$ and $h \simeq 4 \times 10^{-15} \ eV \cdot s$)

The ground state energy of a hydrogen atom is $-13.6 \text{ eV}$. The potential and kinetic energies of the electron in this state are . . . . . . .

The electron of a hydrogen atom makes a transition from the $(n + 1)^{th}$ orbit to the $n^{th}$ orbit. For large $n$,the wavelength of the emitted radiation is proportional to:

An electron in the hydrogen atom jumps from the first excited state to the ground state. What will be the percentage change in the speed of the electron (in $\%$)?