When an \(\alpha\text-\)particle of mass \(m\) moving with velocity \(v\) bombards on a heavy nucleus of charge \(Ze\), its distance of closest approach from the nucleus depends on \(m\) as:
1. \(\frac{1}{\sqrt{m}}\)
2. \(\frac{1}{m^{2}}\)
3. \(m\)
4. \( \frac{1}{m}\)

The energy of a hydrogen atom in the ground state is \(-13.6\) eV. The energy of a He⁺ ion in the first excited state will be:
1. \(-13.6\) eV
2. \(-27.2\) eV
3. \(-54.4\) eV
4. \(-6.8\) eV

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is:
1. \(1\)
2. \(4\)
3. \(0.5\)
4. \(2\)

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

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)