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

The radius of the first orbit of hydrogen is $r_{H}$,and the energy in the ground state is $-13.6 \text{ eV}$. Considering a $\mu^{-}$-particle with a mass $207 m_e$ revolving around a proton as in a hydrogen atom,the energy and radius of the proton and $\mu^{-}$-combination respectively in the first orbit are (assume the nucleus to be stationary):

$A$ particle of mass $m$ moves around the origin in a potential $\frac{1}{2} m \omega^{2} r^{2}$,where $r$ is the distance from the origin. Applying the Bohr's model in this case,the radius of the particle in its $n$th orbit in terms of $a=\sqrt{\frac{h}{2 \pi m \omega}}$ is

Kinetic energy of an electron in one of the orbits of a hydrogen atom is $x$. Then, its total energy is . . . . . . .

When a hydrogen atom emits a photon of energy $12.09 \,eV$,its orbital angular momentum changes by (where $h$ is Planck's constant)

$A$ particular hydrogen-like ion emits radiation of frequency $2.92 \times 10^{15} \text{ Hz}$ when it makes a transition from $n=3$ to $n=1$. The frequency in $\text{Hz}$ of radiation emitted in the transition from $n=2$ to $n=1$ will be: (in $\times 10^{15}$)