The angular momentum of the electron in the third Bohr orbit of hydrogen atom is $l$. Its angular momentum in the fourth Bohr orbit is

The ratio of the area of the first excited state to the ground state orbit of a hydrogen atom is:

An electron makes a transition from orbit $n = 4$ to the orbit $n = 2$ of a hydrogen atom. The wave number of the emitted radiations ($R =$ Rydberg's constant) will be

In the Bohr model of the hydrogen atom,the ratio of the periods of revolution of an electron in $n = 2$ and $n = 1$ orbits is: (in $: 1$)

$A$ 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:

For $He^{+}$, a transition takes place from the orbit of radius $105.8 \ pm$ to the orbit of radius $26.45 \ pm$. The wavelength (in $nm$) of the emitted photon during the transition is. . . . .
[Use: Bohr radius, $a_0=52.9 \ pm$; Rydberg constant, $R_H=2.2 \times 10^{-18} \ J$; Planck's constant, $h=6.6 \times 10^{-34} \ J \ s$; Speed of light, $c=3 \times 10^8 \ m \ s^{-1}$]