The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because:
1. | of the electrons not being subjected to a central force. |
2. | of the electrons colliding with each other. |
3. | of screening effects. |
4. | the force between the nucleus and an electron will no longer be given by Coulomb's law. |
The binding energy of a H-atom, considering an electron moving around a fixed nucleus (proton), is,
\(B = - \dfrac{me^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}\) (\(\mathrm{m}=\) 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,
\(B = - \dfrac{me^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}\) (\(\mathrm{M}=\) proton mass)
This last expression is not correct, because,
1. | \(\mathrm{n}\) would not be integral. |
2. | Bohr-quantisation applies only to electron. |
3. | the frame in which the electron is at rest is not inertial. |
4. | the motion of the proton would not be in circular orbits, even approximately. |
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. \(1.2\times10^6~\text{m/s}\)
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)
1. | \(7.6\times10^{13}\) Hz | 2. | \(4.7\times10^{15}\) Hz |
3. | \(6.6\times10^{15}\) Hz | 4. | \(5.2\times10^{13}\) Hz |
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. \(5.3\times10^{45}\)
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 minimum orbital angular momentum of the electron in a hydrogen atom is:
1. \(h\)
2. \(h/2\)
3. \(h/2\pi\)
4. \(h/ \lambda\)