The transition from the state \(n=3\) to \(n=1\) in hydrogen-like atoms results in ultraviolet radiation. Infrared radiation will be obtained in the transition from:
1. \(3\rightarrow 2\)
2. \(4\rightarrow 2\)
3. \(4\rightarrow 3\)
4. \(2\rightarrow 1\)

In a Geiger-Marsden experiment, what is the distance of the closest approach to the nucleus of a \(7.7\) MeV \(\alpha\)-particle before it comes momentarily to rest and reverses its direction?
1. \(10\) fm
2. \(25\) fm
3. \(30\) fm
4. \(35\) fm

It is found experimentally that \(13.6~\text{eV}\) energy is required to separate a hydrogen atom into a proton and an electron. The velocity of the electron in a hydrogen atom is:
1. \(3.2\times10^6~\text{m/s}\)
2. \(2.2\times10^6~\text{m/s}\)
3. \(3.2\times10^6~\text{m/s}\)
4. \(1.2\times10^6~\text{m/s}\)$\mathrm{}$

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 wavelength of the first spectral line of the Lyman series of the hydrogen spectrum is:
1. \(1218\) Å
2. \(974.3\) Å
3. \(2124\) Å
4. \(2120\) Å

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 simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because:
 

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,

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