For an electron moving in the $n^{th}$ orbit of an $H$-atom,the angular velocity is proportional to:

Let $v_{n}$ and $E_{n}$ be the respective speed and energy of an electron in the $n$th orbit of radius $r_{n}$,in a hydrogen atom,as predicted by Bohr's model. Then:

The frequency of the $1^{st}$ line of the Balmer series in the ${H_2}$ atom is ${\nu _0}$. What is the frequency of the corresponding line emitted by a singly ionized ${He^+}$ atom?

The ratio of radii of $3^{\text{rd}}$ and $6^{\text{th}}$ Bohr's orbit in a hydrogen atom is

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

$A$ small particle of mass $m$ moves in such a way that its potential energy $U = \frac{1}{2} m \omega^2 r^2$,where $\omega$ is a constant and $r$ is the distance of the particle from the origin. Assuming Bohr's quantization of angular momentum and a circular orbit,the radius of the $n^{\text{th}}$ orbit will be proportional to: