Q. No.| Statement I: | The time period of revolution of an electron in its \(n^\mathrm{th}\) Bohr orbit in an atom is directly proportional to \(n^3.\) |
| Statement II: | The kinetic energy of an electron in its \(n^\mathrm{th}\) Bohr orbit in an atom is directly proportional to \(n.\) |
| 1. | Statement I is incorrect and Statement II is correct. |
| 2. | Both Statement I and Statement II are correct. |
| 3. | Both Statement I and Statement II are incorrect. |
| 4. | Statement I is correct and Statement II is incorrect. |
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 |
| 1. | \( n_1 = 6~\text{and}~n_2 = 2\) | 2. | \( n_1 = 8~\text{and}~ n_2 = 1\) |
| 3. | \( n_1 = 8~\text{and}~ n_2 = 2\) | 4. | \(n_1 = 4~\text{and}~n_2 = 2\) |
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. \(5.3\times10^{45}\)
| 1. | \(9:4\) | 2. | \(1:4\) |
| 3. | \(4:1\) | 4. | \(4:9\) |
The minimum orbital angular momentum of the electron in a hydrogen atom is:
1. \(h\)
2. \(h/2\)
3. \(h/2\pi\)
4. \(h/ \lambda\)
| 1. | \(1:2\) | 2. | \(2:1\) |
| 3. | \(3:2\) | 4. | \(2:3\) |
| 1. | \(0.25\) | 2. | \(0.5\) |
| 3. | \(2\) | 4. | \(4\) |
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,
\(B = - \dfrac{me^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}\) (\(\mathrm{m}=\) 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. |
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