Imagine removing one electron from $\mathrm{He}^4$ and $\mathrm{He}^3$. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why?
On removing one electron from $\mathrm{He}^4$ and $\mathrm{He}^3$, the energy levels, as worked out on the basis of Bohr model will be very close as both the nuclei are very heavy as compared to electron mass.Also after removing one electron from $\mathrm{He}^4$ and $\mathrm{He}^3$ atoms contain one electron and are hydrogen like atoms.
When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation. Why cannot it be emitted as other forms of energy?
The transition of an electron from a higher energy to a lower energy level can appears in the form of electromagnetic radiation because electrons interact only electromagnetically.
Would the Bohr formula for the H -atom remain unchanged if proton had a charge $(+4 / 3) e$ and electron a charge $(-3 / 4) e$, where $e=1.6 \times 10^{-19} \mathrm{C}$. Give reasons for your answer.
If proton had a charge $(+4 / 3) e$ and electron a charge $(-3 / 4) e$, then the Bohr formula for the H -atom remain same, since the Bohr formula involves only the product of the charges which remain constant for given values of charges.
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
$$\frac{-13.6 \mathrm{eV}}{2}=-6.8 \mathrm{~eV}$$