Q4. A small particle of mass m moves in such a way that its potential energy U = 12 mω2r2 where ω is constant and r is the distance of the particle from origin. Assuming Bohr’s quantization of momentum and circular orbit, the radius of nth orbit will be proportional to (1) √n (2) n1 (3) n2 (4) n
What This Question Tests
This question combines potential energy, force calculation, uniform circular motion dynamics, and Bohr's quantization condition to find the proportionality of the orbital radius.
Concepts Tested
Formulas Used
F = -dU/dr
F = mv²/r
L = mvr = nħ
U = 1/2 mω²r²
📚 NCERT Sections This Tests
12.7 — The Radius Of The Innermost Electron Orbit Of A Hydrogen Atom Is
Physics Class 12 · Chapter 12
12.7 The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?
12.6 — (A) Using The Bohr’S Model Calculate The Speed Of The Electron In A
Physics Class 12 · Chapter 12
12.6 (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.
12.4 — Bohr Model Of The Hydrogen
Physics Class 12 · Chapter 12
12.4 BOHR MODEL OF THE HYDROGEN ATOM The model of the atom proposed by Rutherford assumes that the atom, consisting of a central nucleus and revolving electron is stable much like sun-planet system which the model imitates. However, there are some fundamental differences between the two situations. While the planetary system is held by gravitationalforce, the nucleus-electron system being charged NIELS objects, interact by Coulomb’s Law of force. We know that an object which moves in a circle is being constantly accelerated – the acceleration being centripetal in nature. According to classical HENRIK electromagnetic theory, an accelerating charged particle emits radiation in the form of electromagnetic waves. The energy of an accelerating electron should therefore, Niels Henrik David Bohrcontinuously decrease. The electron would spiral (1885 – 1962) Danish DAVID inward and eventually fall into the nucleus (Fig. 12.6). physicist who explained the Thus, such an atom can not be stable. Further, spectrum of hydrogen atom according to the classical electromagnetic theory, the based on quantum ideas. BOHR He gave a theory of nuclearfrequency of the electromagnetic waves emitted by the fission based on the liquid- revolving electrons is equal to the frequency of drop model of nucleus. revolution. As the electrons spiral inwards, their angular Bohr contributed to the (1885 velocities and hence their frequencies would change clarification of conceptual – continuously, and so will the frequency of the light problems in quantum emitted. Thus, they would emit a continuous spectrum, mechanics, in particular by in contradiction to the line spectrum actually observed. proposing the comple- 1962) Clearly Rutherford model tells only a part of the story mentary principle. implying that the classical ideas are not sufficient to explain the atomic structure. 297 Reprint 2025-26 Physics FIGURE 12.6 An accelerated atomic electron must spiral into the nucleus as it loses energy. Example 12.4 According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom. Solution From Example 12.3 we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10–11 m is 2.2 × 10–6 m/s. Thus, the frequency of the electron moving around the proton is v 2.2 × 10 6 m s −1 ν = = −11 2 π r 2 π 5.3 × 10 m ( ) 12.4 » 6.6 × 1015 Hz. According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. EXAMPLE Thus the initial frequency of the light emitted is 6.6 × 1015 Hz. It was Niels Bohr (1885 – 1962) who made certain modifications in this model by adding the ideas of the newly developing quantum hypothesis. Niels Bohr studied in Rutherford’s laboratory for several months in 1912 and he was convinced about the validity of Rutherford nuclear model. Faced with the dilemma as discussed above, Bohr, in 1913, concluded that in spite of the success of electromagnetic theory in explaining large-scale phenomena, it could not be applied to the processes at the atomic scale. It became clear that a fairly radical departure from the established principles of classical mechanics and electromagnetism would be needed to understand the structure of atoms and the relation of atomic structure to atomic spectra. Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates. These are : (i) Bohr’s first postulate was that an electron in an atom could revolve in certain stable orbits without the emission of radiant energy, contrary to the predictions of electromagnetic theory. According to 298 this postulate, each atom has certain definite stable states in which it Reprint 2025-26 Atoms can exist, and each possible state has definite total energy. These are called the stationary states of the atom. (ii) Bohr’s second postulate defines these stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π where h is the Planck’s constant (= 6.6 × 10–34 J s). Thus the angular momentum (L) of the orbiting electron is quantised. That is L = nh/2π (12.5) (iii) Bohr’s third postulate incorporated into atomic theory the early quantum concepts that had been developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by hν = Ei – Ef (12.6) where Ei and Ef are the energies of the initial and final states and Ei > Ef. For a hydrogen atom, Eq. (12.4) gives the expression to determine the energies of different energy states. But then this equation requires the radius r of the electron orbit. To calculate r, Bohr’s second postulate about the angular momentum of the electron–the quantisation condition – is used. The radius of nth possible orbit thus found is n 2 h 2 4 πε0 rn = 2 (12.7) m 2 π e The total energy of the electron in the stationary states of the hydrogen atom can be obtained by substituting the value of orbital radius in Eq. (12.4) as e 2 m 2 π 2 e 2 E n = − 2 8 πε0 n h 4 πε0 me 4 or E n = − 2 2 2 (12.8) 8n ε0 h Substituting values, Eq. (12.8) yields −18 2.18 × 10 E n = − 2 J (12.9) n Atomic energies are often expressed in electron volts (eV) rather than joules. Since 1 eV = 1.6 × 10–19 J, Eq. (12.9) can be rewritten as 13.6 E n = − 2 eV (12.10) n The negative sign of the total energy of an electron moving in an orbit means that the electron is bound with the nucleus. Energy will thus be required to remove the electron from the hydrogen atom to a distance 299 infinitely far away from its nucleus (or proton in hydrogen atom). Reprint 2025-26 Physics 12.4.1 Energy levels The energy of an atom is the least (largest negative value) when its electron is revolving in an orbit closest to the nucleus i.e., the one for which n = 1. For n = 2, 3, ... the absolute value of the energy E is smaller, hence the energy is progressively larger in the outer orbits. The lowest state of the atom, called the ground state, is that of the lowest energy, with the electron revolving in the orbit of smallest radius, the Bohr radius, a 0. The energy of this state (n = 1), E1 is –13.6 eV. Therefore, the minimum energy required to free the electron from the ground state of the hydrogen atom is 13.6 eV. It is called the ionisation energy of the hydrogen atom. This prediction of the Bohr’s model is in excellent agreement with the experimental value of ionisation energy. At room temperature, most of the hydrogen atoms are in ground state. When a hydrogen atom receives energy by processes such as electron collisions, the atom may acquire sufficient energy to raise the electron to higher energy states. The atom is then said to be in an excited state. From Eq. (12.10), for n = 2; the energy E2 is –3.40 eV. It means that the energy required to excite an electron in hydrogen atom to its first excited state, is an FIGURE 12.7 The energy level energy equal to E2 – E1 = –3.40 eV – (–13.6) eV = 10.2 eV. diagram for the hydrogen atom. Similarly, E3 = –1.51 eV and E3 – E1 = 12.09 eV, or to exciteThe electron in a hydrogen atom the hydrogen atom from its ground state (n = 1) to second at room temperature spends most of its time in the ground excited state (n = 3), 12.09 eV energy is required, and so state. To ionise a hydrogen on. From these excited states the electron can then fall back atom an electron from the to a state of lower energy, emitting a photon in the process. ground state, 13.6 eV of energy Thus, as the excitation of hydrogen atom increases (that is must be supplied. (The horizontal as n increases) the value of minimum energy required to lines specify the presence of free the electron from the excited atom decreases. allowed energy states.) The energy level diagram* for the stationary states of a hydrogen atom, computed from Eq. (12.10), is given in Fig. 12.7. The principal quantum number n labels the stationary states in the ascending order of energy. In this diagram, the highest energy state corresponds to n =¥ in Eq, (12.10) and has an energy of 0 eV. This is the energy of the atom when the electron is completely removed (r = ¥) from the nucleus and is at rest. Observe how the energies of the excited states come closer and closer together as n increases. 12.5 THE LINE SPECTRA OF THE HYDROGEN ATOM According to the third postulate of Bohr’s model, when an atom makes a transition from the higher energy state with quantum number ni to the lower energy state with quantum number nf (nf < ni), the difference of energy is carried away by a photon of frequency nif such that * An electron can have any total energy above E = 0 eV. In such situations the 300 electron is free. Thus there is a continuum of energy states above E = 0 eV, as shown in Fig. 12.7. Reprint 2025-26 Atoms hvif = Eni – Enf (12.11) Since both nf and ni are integers, this immediately shows that in transitions between different atomic levels, light is radiated in various discrete frequencies. The various lines in the atomic spectra are produced when electrons jump from higher energy state to a lower energy state and photons are emitted. These spectral lines are called emission lines. But when an atom absorbs a photon that has precisely the same energy needed by the electron in a lower energy state to make transitions to a higher energy state, the process is called absorption. Thus if photons with a continuous range of frequencies pass through a rarefied gas and then are analysed with a spectrometer, a series of dark spectral absorption lines appear in the continuous spectrum. The dark lines indicate the frequencies that have been absorbed by the atoms of the gas. The explanation of the hydrogen atom spectrum provided by Bohr’s model was a brilliant achievement, which greatly stimulated progress towards the modern quantum theory. In 1922, Bohr was awarded Nobel Prize in Physics.
📋 Question Details
- Chapter
- Atoms
- Topic
- Bohr's quantization condition
- Year
- 2023
- Shift
- 06 Apr Shift 2
- Q Number
- Q4
- Type
- MCQ
- NCERT Ref
- Class 12 Physics Ch 12: Atoms
More from this Chapter
Q70.If 13.6eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is (1) 10.2eV (2) 0eV (3) 3.4eV (4) 6.8eV
Q79.In a hydrogen atom, if energy of an electron in ground state is 13.6 eV, then that in the 2nd excited state is JEE Main 2002 JEE Main Previous Year Paper (1) 1.51 eV (2) 3.4 eV (3) 6.04 eV (4) 13.6 eV
Q69.The wavelengths involved in the spectrum of deuterium (21D) are slightly different from that of hydrogen spectrum, because (1) the size of the two nuclei are different (2) the nuclear forces are different in the two cases (3) the masses of the two nuclei are different (4) the attraction between the electron and the nucleus is different in the two cases
Q70.If the binding energy of the electron in a hydrogen atom is 13.6eV, the energy required to remove the electron from the first excited state of Li++ is (1) 30.6eV (2) 13.6eV (3) 3.4eV (4) 122.4eV