Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?
According to Bohr model electrons having different energies belong to different levels having different values of $n$. So, their angular momenta will be different, as
$$L=\frac{n h}{2 \pi} \text { or } L \propto n$$
Positronium is just like a H -atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?
The total energy of the electron in the stationary states of the hydrogen atom is given by
$$E_n=-\frac{m e^4}{8 n^2 \varepsilon_0^2 h^2}$$
where signs are as usual and the $m$ that occurs in the Bohr formula is the reduced mass of electron and proton. Also, the total energy of the electron in the ground state of the hydrogen atom is -13.6 eV . For H -atom reduced mass $m_e$. Whereas for positronium, the reduced mass is
$$m \approx \frac{m_e}{2}$$
Hence, the total energy of the electron in the ground state of the positronium atom is
$$\frac{-13.6 \mathrm{eV}}{2}=-6.8 \mathrm{~eV}$$
Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb's law as usual. Under such circumstances, calculate the ground state energy of a He-atom.
For a He -nucleus with charge $2 e$ and electrons of charge $-e$, the energy level in ground state is
$$-E_n=Z^2 \frac{-13.6 \mathrm{eV}}{n^2}=2^2 \frac{-13.6 \mathrm{eV}}{1^2}=-54.4 \mathrm{eV}$$
Thus, the ground state will have two electrons each of energy $E$ and the total ground state energy would be $-(4 \times 13.6) \mathrm{eV}=-54.4 \mathrm{eV}$.
Using Bohr model, calculate the electric current created by the electron when the H -atom is in the ground state.
The electron in Hydrogen atom in ground state revolves on a circular path whose radius is equal to the Bohr radius $\left(a_n\right)$.
Let the velocity of electron is $v$.
$\therefore \quad$ Number of revolutions per unit time $=\frac{2 \pi a_0}{v}$
The electric current is given by $i=\frac{q}{t}$, if $q$ charge flows in timet. Here, $q=e$ The electric current is given by $i=\frac{2 \pi a_0}{v} e$.
Show that the first few frequencies of light that is emitted when electrons fall to $n$th level from levels higher than $n$, arc approximate harmonics (i.e., in the ratio $1: 2: 3 \ldots$ ) when $n>>1$.
The frequency of any line in a series in the spectrum of hydrogen like atoms corresponding to the transition of electrons from $(n+p)$ level to $n$th level can be expressed as a difference of two terms;
$$v_{m n}=c R Z^2\left[\frac{1}{(n+p)^2}-\frac{1}{n^2}\right]$$
where, $m=n+p,(p=1,2,3, \ldots)$ and $R$ is Rydberg constant.
$\begin{aligned} &\begin{aligned} \text{For}\quad p & \ll n \\ v_{m n} & =c R Z^2\left[\frac{1}{n^2}\left(1+\frac{p}{n}\right)^{-2}-\frac{1}{n^2}\right] \\ v_{m n} & =c R Z^2\left[\frac{1}{n^2}-\frac{2 p}{n^3}-\frac{1}{n^2}\right] \end{aligned}\\ &\text { [By binomial theorem } \left.(1+x)^n=1+n x \text { if }|x|<1\right]\\ &v_{m n}=c R Z^2 \frac{2 p}{n^3} \simeq\left(\frac{2 c R Z^2}{n^3}\right) p \end{aligned}$
Thus, the first few frequencies of light that is emitted when electrons fall to the $n$th level from levels higher than $n$, are approximate harmonic (i.e., in the ratio $1: 2: 3 \ldots$ ) when $n \gg 1$.