The magnetic moment of an electron due to its orbital motion is proportional to $(n =$ principal quantum number$)$

The orbital magnetic moment $(m_{orb})$ of a revolving electron around the nucleus varies with the principal quantum number $(n)$ as

$A$ particle of mass $200 \, MeV/c^2$ collides with a hydrogen atom at rest. Soon after the collision,the particle comes to rest,and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in $eV$) is $\frac{N}{4}$. The value of $N$ is: (Given the mass of the hydrogen atom to be $1 \, GeV/c^2$)

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?

Let $R_1$ be the radius of the second stationary orbit and $R_2$ be the radius of the fourth stationary orbit of an electron in Bohr's model. The ratio $\frac{R_1}{R_2}$ is

Which state of triply ionised Beryllium $(Be^{+++})$ has the same orbital radius as that of the ground state of hydrogen?