Q. No.For the ground state, the electron in the H-atom has an angular momentum \(\dfrac h{2\pi}\), according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,
1.
because the Bohr model gives incorrect values of angular momentum.
2.
because only one of these would have a minimum energy.
3.
angular momentum must be in the direction of the spin of the electron.
4.
because electrons go around only in horizontal orbits.
A set of atoms in an excited state decays:
| 1. | in general to any of the states with lower energy. |
| 2. | into a lower state only when excited by an external electric field. |
| 3. | all together simultaneously into a lower state. |
| 4. | to emit photons only when they collide. |
An ionised \(\text H\)-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state:
| (a) | the electron would not move in circular orbits. |
| (b) | the energy would be \(2^{4}\) times that of a \(\text H\)-atom. |
| (c) | the electron's orbit would go around the protons. |
| (d) | the molecule will soon decay in a proton and a \(\text H\)-atom. |
Choose the correct option:
| 1. | (a), (b) | 2. | (a), (c) |
| 3. | (b), (c), (d) | 4. | (c), (d) |
Let \(E_{n} = \frac{- 1m e^{4}}{8 \varepsilon_{0}^{2}n^{2} h^{2}} \) be the energy of the \(n^\text{th}\) level of H-atom. If all the H-atoms are in the ground state and radiation of frequency \(\frac{\left(\right. E_{2} - E_{1} \left.\right)}{h}\) falls on it, then:
| (a) | it will not be absorbed at all. |
| (b) | some of the atoms will move to the first excited state. |
| (c) | all atoms will be excited to the \(n = 2\) state. |
| (d) | no atoms will make a transition to the \(n = 3\) state. |
Choose the correct option:
| 1. | (b, d) | 2. | (a, d) |
| 3. | (b, c, d) | 4. | (c, d) |
The simple Bohr model is not applicable to \(\text{He}^4\) atom because:
| (a) | \(\text{He}^4\) is an inert gas. |
| (b) | \(\text{He}^4\) has neutrons in the nucleus. |
| (c) | \(\text{He}^4\) has one more electron. |
| (d) | electrons are not subject to central forces. |
Choose the correct option:
| 1. | (a), (c) | 2. | (a), (c), (d) |
| 3. | (b), (d) | 4. | (c), (d) |
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. |
A hydrogen ion and a singly ionised helium atom are accelerated from rest through the same potential difference. The ratio of their final speeds is close to:
| 1. | \(1:2\) | 2. | \(10:7\) |
| 3. | \(5:7\) | 4. | \(2:1\) |
Assume that an electron orbits around a nucleus in a circular orbit, and Newton's laws are valid. What is the ratio of the kinetic energy \((KE)\) of the electron to its potential energy \((PE)\) in orbit (in magnitude)?
1. \(1:2\)
2. \(2:1\)
3. \(1:1\)
4. none of the above
| 1. | \(13.6\) eV | 2. | \(\dfrac{13.6} {2}\) eV |
| 3. | \(2 \times 13.6\) eV | 4. | \(10.2\) eV |
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