In a hydrogen atom,an electron excites from the ground state to a higher energy state,and its orbital velocity is reduced to $\frac{1}{3}$ of its initial value. The radius of the orbit in the ground state is $R$. The radius of the orbit in that higher energy state is: (in $R$)

$A$ moving Hydrogen atom makes a head-on collision with a stationary Hydrogen atom. Before the collision,both atoms are in the ground state,and after the collision,they move together. The minimum kinetic energy of the moving Hydrogen atom,such that one of the atoms reaches the excitation state,is (in $eV$)

Ratio of centripetal acceleration for an electron revolving in $3^{\text{rd}}$ and $5^{\text{th}}$ Bohr orbit of hydrogen atom is

$A$ 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:

In $(i)$ a hydrogen atom and $(ii)$ a singly ionized helium atom,an electron makes a transition from the same excited state $n$ to the ground state. The ratio of the wavelengths of the emitted photons in the two cases will be:

Consider a hydrogen-like atom whose energy in the $n^{th}$ excited state is given by $E_n = - \frac{13.6 Z^2}{n^2}$. When this excited atom makes a transition from an excited state to the ground state, the most energetic photons have energy $E_{max} = 52.224 \ eV$ and the least energetic photons have energy $E_{min} = 1.224 \ eV$. The atomic number of the atom is: