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

In a hydrogen atom and a $Li^{2+}$ ion,the electron is in the second excited state. If $l_{H}$ and $l_{Li}$ are the angular momenta of the electrons and $E_H$ and $E_{Li}$ are their respective energies,then:

The Bohr model for the $H$-atom relies on Coulomb's law of electrostatics. Coulomb's law has not been directly verified for very short distances of the order of $\mathring{A}$. Suppose Coulomb's law between two opposite charges $+q_1$ and $-q_2$ is modified to $|\vec{F}| = \frac{q_1 q_2}{4\pi \epsilon_0} \left( \frac{1}{r^2} \right)$ for $r \ge R_0$ and $|\vec{F}| = \frac{q_1 q_2}{4\pi \epsilon_0} \left( \frac{1}{R_0^{2-\epsilon} r^{\epsilon}} \right)$ for $r < R_0$. Calculate the ground state energy of an $H$-atom,given $\epsilon = 0.1$ and $R_0 = 1 \,\mathring{A}$.

In the following atoms and molecules,for the transition from $n = 2$ to $n = 1$,the spectral line of minimum wavelength will be produced by:

Show that the first few frequencies of light that is emitted when electrons fall to the $n^{\text{th}}$ level from levels higher than $n$,are approximate harmonics (i.e.,in the ratio $1:2:3...$) when $n >> 1$.

An electron in a hydrogen atom,after absorbing energy photons,can jump between energy states $n_1$ and $n_2$ $(n_2 > n_1)$. It then returns to the ground state,emitting six different wavelengths in the emission spectrum. The energy of the emitted photons can be equal to,less than,or greater than the absorbed photon energy. Determine $n_1$ and $n_2$.