Q. No.Which statement about the Rutherford model of the atom is not true?
| 1. | there is a positively charged centre in an atom called the nucleus. |
| 2. | nearly all the mass of an atom resides in the nucleus. |
| 3. | the size of the nucleus is the same as that of the atom. |
| 4. | electrons occupy the space surrounding the nucleus. |
| Statement I: | For the scattering of \(\alpha\) -particles at large angles, only the nucleus of the atom is responsible. |
| Statement II: | The nucleus is very heavy in comparison to electrons. |
| 1. | Statement I is correct and Statement II is incorrect. |
| 2. | Statement I is incorrect and Statement II is correct. |
| 3. | Both Statement I and Statement II are correct. |
| 4. | Both Statement I and Statement II are incorrect. |
In \(1911\), the physician Ernest Rutherford discovered that atoms have a tiny, dense nucleus by shooting positively charged particles at a very thin gold foil. A key physical property that led Rutherford to use gold was that it was:
1. electrically conducting
2. highly malleable
3. shiny
4. non-reactive
| 1. | \(2\) protons only. |
| 2. | \(2\) protons and \(2\) neutrons only. |
| 3. | \(2\) electrons, \(2\) protons, and \(2\) neutrons. |
| 4. | \(2\) electrons and \(4\) protons only. |
Which of the following curves may represent a variation of no. of \(\alpha-\)particles scattered (\(N\)) with scattering angle (\(\theta\)) in Rutherford's \(\alpha-\)particle scattering experiment?
| 1. |  |
2. |  |
| 3. |  |
4. |  |
| Assertion (A): | The positively charged nucleus of an atom has a radius of almost \(10^{-15}~\text{m}\). |
| Reason (R): | In \(\alpha\)-particle scattering experiment, the distance of the closest approach for \(\alpha\)-particle is \(\approx 10^{-15}~\text m\). |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | Both (A) and (R) are False. |
In Rutherford’s nuclear model of the atom, the nucleus (radius about \(10^{-15}~\text{m}\)) is analogous to the sun about which the electron move in orbit (radius \(\approx 10^{-10}~\text{m}\)) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, then: (The radius of the earth's orbit is about \(1.5\times 10^{11}~\text{m}\). The radius of the sun is taken as \(7\times10^{8}~\text{m}\).)
| 1. | the earth will be closer to the sun than it is actually. |
| 2. | the earth will be farther away from the sun than it is actually. |
| 3. | the earth remains at the same distance from the sun as it is actually. |
| 4. | None of these |
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
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. |
1. \(0\)
2. \(1\)
3. \(2\)
4. \(3\)
© 2024 GoodEd Technologies Pvt. Ltd.