The simple Bohr model is not applicable to $\mathrm{He}^4$ atom because
The mass of a H -atom is less than the sum of the masses of a proton and electron. Why is this?
Since, the difference in mass of a nucleus and its constituents, $\Delta M$, is called the mass defect and is given by
$$\Delta M=\left[Z m_p+(A-Z) m_n\right]-M$$
Also, the binding energy is given by $B=$ mass defect $(\Delta M) \times c^2$.
Thus, the mass of a H -atom is $m_p+m_e-\frac{B}{c^2}$, where $B \approx 13.6 \mathrm{eV}$ is the binding energy.
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.