If $m$ is the mass of an electron,$v$ is its velocity,$r$ is the radius of a stationary circular orbit around a nucleus with charge $Ze$,then from Bohr's first postulate,the kinetic energy of the electron is (where $K = 1 / 4 \pi \epsilon_0$):

The emission series of hydrogen atom is given by $\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$ where,$R$ is the Rydberg constant. For a transition from $n_{2}$ to $n_{1}$,the relative change $\Delta \lambda / \lambda$ in the emission wavelength,if hydrogen is replaced by deuterium (assume that,the mass of proton and neutron are the same and approximately $2000$ times larger than that of electrons) is ........... $\%$

In Bohr's model,the atomic radius of the first orbit is $r_0$,then the radius of the third orbit is

$A$ muon is an unstable particle with a mass of $207 \, m_e$ and a charge of either $+e$ or $-e$. $A$ muon $(\mu^-)$ is captured by a hydrogen nucleus to form a muonic atom. If the proton captures the $\mu^-$, find the ionization energy of this atom in $keV$.

The ground state energy of a hydrogen atom is $-13.6 \text{ eV}$. What is the ratio of the kinetic energy to the potential energy of the electron in this state?

An orbital electron in the ground state of hydrogen has a magnetic moment $\mu_1$. This orbital electron is excited to the $3^{rd}$ excited state by some energy transfer to the hydrogen atom. If the new magnetic moment of the electron is $\mu_2$,then: