According to the classical electromagnetic theory, the initial frequency of the light emitted by the electron revolving around a proton in the hydrogen atom is: (The velocity of the electron moving around a proton in a hydrogen atom is \(2.2\times10^{6}\) m/s)

A \(10~\text{kg}\) satellite circles earth once every \(2~\text{h}\) in an orbit having a radius of \(8000~\text{km}\). Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom. The quantum number of the orbit of the satellite is:
1. \(2.0\times10^{43}\)
2. \(4.7\times10^{45}\)
3. \(3.0\times10^{43}\)
4. \(5.3\times10^{45}\)$\mathrm{}$

The minimum orbital angular momentum of the electron in a hydrogen atom is:
1. \(h\)
2. \(h/2\)
3. \(h/2\pi\)
4. \(h/ \lambda\)

Let \(L_1\) and \(L_2\) be the orbital angular momentum of an electron in the first and second excited states of the hydrogen atom, respectively. According to Bohr's model, the ratio \(L_1:L_2\) is:
1. \(1:2\)
2. \(2:1\)
3. \(3:2\)
4. \(2:3\)

Taking the bohr radius as \(a_0=53\) pm, the radius of Li⁺⁺ ion in its ground state on the basis of bohr's model will be about:
1. \(153\) pm
2. \(27\) pm
3. \(18\) pm
4. \(13\) pm

The binding energy of a H-atom, considering an electron moving around a fixed nucleus (proton), is,

$\mathrm{B}=-\frac{{\mathrm{me}}^4}{8{\mathrm{n}}^2{\mathrm{\epsilon }}_0^2{\mathrm{h}}^2}$ (\(\mathrm{m}=\) electron mass)
If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be,

$\mathrm{B}=-\frac{{\mathrm{Me}}^4}{8{\mathrm{n}}^2{\mathrm{\epsilon }}_0^2{\mathrm{h}}^2}$ (\(\mathrm{M}=\) proton mass)
This last expression is not correct, because,