The radius of a hydrogen atom in its ground state is $5.3 \times 10^{-11} \ m$. After collision with an electron,it is found to have a radius of $21.2 \times 10^{-11} \ m$. What is the principal quantum number $n$ of the final state of the atom?

How is the linear velocity $v$ of an electron in a $Bohr$ orbit related to its principal quantum number $n$?

In the Bohr model,an electron moves in a circular orbit around the nucleus. Considering an orbiting electron to be a circular current loop,the magnetic moment of the hydrogen atom,when the electron is in the $n^{th}$ excited state,is ($e=$ electronic charge,$m_{e}=$ mass of the electron,$h=$ Planck's constant).

$A$ hydrogen ion $(H^+)$ and a singly ionized helium atom $(He^+)$ are accelerated from rest through the same potential difference $(V)$. The ratio of the final speeds of the hydrogen ion and the helium ion is close to:

The radius of orbit of an electron and the speed of an electron in the ground state of a hydrogen atom are $5.5 \times 10^{-11} \,m$ and $4 \times 10^6 \,m/s$ respectively. Then,the orbital period of this electron in the first excited state will be

The radius of a hydrogen atom in its ground state is $5.3 \times 10^{-11} \ m$. After collision with an electron,it is found to have a radius of $21.2 \times 10^{-11} \ m$. What is the principal quantum number $n$ of the final state of the atom?