NCERT Exemplars Atoms - Physics Class 12 - NEET PDF Download

MULTIPLE CHOICE QUESTIONS

Q.1. Taking the Bohr radius as a₀ = 53 pm, the radius of Li⁺⁺ ion in its ground state, on the basis of Bohr's model, will be about 
(a) 53 pm 
(b) 27 pm 
(c) 18 pm 
(d) 13 pm
Ans: (c)
Explanation: For a hydrogen-like ion, Bohr's formula gives r_n = a₀ (n²/Z). For the ground state n = 1. Lithium nucleus has Z = 3, so r = a₀/3 = 53/3 ≈ 17.7 pm ≈ 18 pm. The images above show the derivation and the relation r_n ∝ 1/Z.

Q.2. The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), 

 (m = electron mass).

If one decides to work in a frame of reference where the electron is at rest, the proton would be moving arround it. By similar arguments, the binding energy would be

This last expression is not correct because 

(a) n would not be integral. 

(b) Bohr - quantisation applies only to electron 

(c) The frame in which the electron is at rest is not inertial. 

(d) The motion of the proton would not be in circular orbits, even approximately.

Ans: (c)

Explanation: The frame in which the electron is held at rest would be accelerating (it is not an inertial frame). Bohr's quantisation conditions are imposed in an inertial frame where centripetal and Coulomb forces balance in the usual way. Changing to a non-inertial frame alters the forces and the simple Bohr derivation is invalid in that frame. The correct treatment uses the reduced mass of the electron-proton system in an inertial centre-of-mass frame (shown in the image placeholders).

Q.3. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because 
(a) Of the electrons not being subject to a central force
(b) Of the electrons colliding with each other 
(c) Of screening effects 
(d) The force between the nucleus and an electron will no longer be given by Coulomb's law
Ans: (a)
Explanation: The Bohr model assumes each electron feels only the Coulomb attraction from the nucleus (a central force). In multi-electron atoms other electrons produce additional forces (repulsion and screening), so an individual electron is not subject to a purely central Coulomb force from the nucleus alone. Hence the simple Bohr model is not directly applicable to many-electron atoms (screening and electron-electron interactions must be included by more advanced methods).

Q.4. For the ground state, the electron in the H-atom has an angular momentum = ћ, 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, 
(a) Because Bohr model gives incorrect values of angular momentum
(b) Because only one of these would have a minimum energy
(c) Angular momentum must be in the direction of spin of electron
(d) Because electrons go around only in horizontal orbits
Ans: (a)
Explanation: The Bohr model treats angular momentum as a scalar magnitude L = nћ and does not address its directional (vector) nature. Quantum theory shows angular momentum is quantised differently (components and magnitude follow different rules). Thus the Bohr model's treatment of angular momentum is not correct in the vector sense and cannot account for the allowed orientations given by modern quantum mechanics. The Bohr value for the magnitude is a semiclassical result but the model cannot describe directional quantisation correctly.

Q.5. O₂ molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms
(a) is not important because nuclear forces are short-ranged.
(b) is as important as electrostatic force for binding the two atoms.
(c) cancels the repulsive electrostatic force between the nuclei.
(d) is not important because oxygen nucleus have equal number of neutrons and protons
Ans: (a)
Explanation: Nuclear forces act only over distances comparable to nuclear sizes (fm scale). Atomic separations in O₂ are of the order of ångströms (10⁵-10⁶ times larger), so nuclear forces are negligible for binding between atoms. Chemical bonding is governed by electrostatic and quantum electronic interactions, not by the short-range nuclear force.

Q.6. Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is 
(a) 10.20 eV 
(b) 20.40 eV 
(c) 13.6 eV 
(d) 27.2 eV
Ans: (a)
Explanation: Initial total energy of two ground-state H atoms = 2 × (-13.6 eV) = -27.2 eV. The largest possible conversion of kinetic energy into internal energy occurs if one atom is excited to n = 2 (energy -3.4 eV) while the other remains in ground state. Final total energy then = -13.6 + (-3.4) = -17.0 eV. The reduction in kinetic energy (magnitude) = (-27.2) - (-17.0) = -10.2 eV, so the maximum decrease is 10.20 eV.

Q.7. A set of atoms in an excited state decays. 
(a) in general to any of the states with lower energy
(b) into a lower state only when excited by an external electric field
(c) all together simultaneously into a lower state 
(d) to emit photons only when they collide
Ans: (a)
Explanation: Excited atoms may decay spontaneously to any lower permitted energy level, subject to selection rules. The decay need not be triggered by an external field, nor do all atoms decay simultaneously; individual atoms decay probabilistically with characteristic lifetimes.

Q.8. An ionised 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 H-atom
(c) The electrons, orbit would go around the protons 
(d) The molecule will soon decay in a proton and a H-atom
Ans: (a, c)
Explanation: In H₂⁺ (one electron, two protons) the electron feels attraction from two centres and does not follow a simple single-centre circular Bohr orbit - it occupies a molecular orbital that extends over both protons. Thus (a) and (c) are correct. Option (b) is not applicable and (d) is not generally true for the stable molecular ion ground state.

Q.9. Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom, 
(a) Because of energy conservation. 
(b) Without simultaneously releasing energy in the from of radiation. 
(c) Because of momentum conservation. 
(d) Because of angular momentum conservation.
Ans: (a, b)
Explanation: Forming a bound H-atom from free electron and proton requires removal of the binding energy. Energy and momentum conservation cannot both be satisfied if no additional particle carries away the excess energy and momentum. Emission of a photon (radiation) during capture provides the required energy and momentum transfer, so without emitting radiation formation is not possible. (Momentum conservation alone does not forbid capture provided radiation is emitted.

Q.10. The Bohr model for the spectra of a H-atom 
(a) Will not be applicable to hydrogen in the molecular from
(b) Will not be applicable as it is for a He-atom
(c) Is valid only at room temperature
(d) Predicts continuous as well as discrete spectral lines
Ans: (a, b)
Explanation: The Bohr model applies to hydrogen-like (single-electron) systems. It does not apply directly to molecular hydrogen (which is a multi-centre, multi-electron problem) nor to neutral helium (which has two electrons causing electron-electron interaction). The model also predicts discrete spectral lines (not continuous) for allowed transitions; temperature alone does not determine applicability.

Q.11. The Balmer series for the H-atom can be observed
(a) If we measure the frequencies of light emitted when an excited atom falls to the ground state
(b) If we measure the frequencies of light emitted due to transitions between excited states and the first excited state
(c) In any transition in a H-atom
(d) As a sequence of frequencies with the higher frequencies getting closely packed
Ans: (b, d)
Explanation: The Balmer series consists of transitions that end at n = 2 (the first excited state). Thus it is observed when electrons fall to n = 2 from higher levels (option b). As the initial level becomes very large the emitted frequencies approach a limit and lines become more closely spaced, so (d) is also correct. Option (a) refers to transitions to the ground state (Lyman series), not Balmer.

Q.12. Let 

 be the energy of the nth level of H-atom. If all the H-atoms are in the ground state and radiation of frequency (E₂ - E₁)/h falls on it,
(a) It will not be absorbed at all
(b) Some of 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

Ans: (b, d)
Explanation: Radiation with energy equal to E₂ - E₁ (hf) can be absorbed by those atoms that make the resonance transition 1 → 2, so some atoms will be excited (b). The photon energy is insufficient for a direct 1 → 3 transition, so no atoms will be promoted to n = 3 by that radiation (d). Not all atoms necessarily absorb the radiation (so (c) is incorrect).

Q.13. The simple Bohr modle is not applicable to He⁴ atom because
(a) He⁴ is an inert gas.
(b) He⁴ has neutrons in the nucleus.
(c) He⁴ has one more electron.
(d) Electrons are not subject to central forces
Ans: (c, d)
Explanation: He⁴ has two electrons, so the single-electron Bohr model cannot account for electron-electron repulsion (option c). In such atoms electrons are not moving under a simple single-centre Coulomb force alone (option d). The presence of neutrons or inert behaviour is not the reason the Bohr single-electron model fails.

VERY SHORT ANSWER TYPE QUESTIONS

Q.14. The mass of a H-atom is less than the sum of the masses of a proton and electron. Why is this?
Ans: When a proton and an electron combine to form a hydrogen atom energy is released (the binding energy). By Einstein's relation E = Δm c², the released energy corresponds to a loss of mass Δm. Thus the mass of the bound atom (M) is less than the sum of the separate masses: Δm = (m_p + m_e) - M. For the hydrogen atom the binding energy is 13.6 eV so the mass defect is small but non-zero. The image placeholders above show the formulae used.

Q.15. Imagine removing one electron from He⁴ and He³. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.
Ans: After removal of one electron both He⁴ and He³ become hydrogen-like (He⁺) with a single electron orbiting a nucleus of charge +2e. The Bohr energy levels depend primarily on the nuclear charge and the reduced mass. The difference in nuclear masses between He⁴ and He³ affects the reduced mass only slightly, so their energy levels computed by Bohr's formula will be very similar.

Q.16. When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation. Why cannot it be emitted as other forms of energy?
Ans: Electrons are charged particles. A change in their motion (acceleration) produces electromagnetic radiation; hence the energy difference appears naturally as photons. Non-radiative channels (for example transfer to other particles) require interactions or collisions; in an isolated atom the allowed and most direct process is emission of electromagnetic radiation.

Q.17. Would the Bohr formula for the H-atom remain unchanged if proton had a charge (+4/3)e and electron a charge (-3/4)e, where e = 1.6 × 10^-19C. Give reasons for your answer.
Ans: Yes. The Coulomb force (and the Bohr formula) depends on the product of the charges q₁q₂. Here q_p × q_e = (+4/3 e) × (-3/4 e) = -e², the same as the usual case. Since the Bohr expressions involve this product, the formulae for radii and energies remain unchanged.

Q.18. Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?
Ans: According to Bohr, the magnitude of orbital angular momentum is L = nћ. Different energies correspond to different principal quantum numbers n, so the angular momentum values would be different. Therefore, in Bohr's model two electrons with different energies cannot have the same orbital angular momentum.

SHORT ANSWER TYPE QUESTIONS

Q.19. Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?
Ans: For positronium the electron and positron have equal masses m_e. The reduced mass μ = m_e m_e/(m_e + m_e) = m_e/2. 

Q.20. Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb's law as usual. Under such circumstances, calculate the ground state energy of a He-atom.
Ans: If electron-electron repulsion is neglected, each electron in He (Z = 2) behaves like a hydrogen-like electron with nuclear charge Z = 2. The energy for n = 1 is E = -13.6 eV × Z² = -13.6 × 4 = -54.4 eV for one electron. If both electrons occupy the same lowest level in this hypothetical case, the total ground-state energy would be 2 × (-54.4 eV) = -108.8 eV.

Q.21. Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.
Ans: The current produced by a revolving charge equals charge × number of revolutions per second.
Radius (ground state) r = a₀ = 5.29 × 10^-11 m.
Bohr gives mvr = ħ ⇒ v = ħ/(m r).
Revolutions per second f = v/(2π r) = ħ/(2π m r²).
Current I = e f = e ħ/(2π m r²).
Using r = a₀, this evaluates numerically to about I ≈ 1.06 × 10^-3 A ≈ 1.06 mA (direction opposite to electron motion, so sign is negative if direction is defined along motion).
 

Q.22. Show that the first few frequencies of light that is emitted when electrons fall to the nth level from levels higher than n, are approximate harmonics (i.e. in the ratio 1 : 2: 3...) when n >>1.
Ans: The frequency for a transition m → n is given by ν = R c (1/n² - 1/m²) with m = n + p (p = 1,2,3...). For large n, expand 1/(n+p)² ≈ 1/n² (1 - 2p/n). Then the difference becomes ν ≈ R c (2p / n³). Thus ν ∝ p for fixed large n, so the first few frequencies (p = 1,2,3...) are approximately in the ratio 1 : 2 : 3. The image placeholders above contain the binomial expansion and intermediate steps.

Q.23. What is the minimum energy that must be given to a H atom in ground state so that it can emit an H_γ line in Balmer series. If the angular momentum of the system is conserved, what would be the angular momentum of such H_γ photon?
Ans: H_γ corresponds to the transition 5 → 2. To produce this emission starting from the ground state the electron must first be excited from n = 1 to n = 5. Minimum energy required = E₅ - E₁ = -13.6/5² - (-13.6) = 13.6(1 - 1/25) = 13.6 × 24/25 = 13.056 eV.
If angular momentum is treated in the Bohr picture, L = nћ, so the change in orbital angular momentum for the 5 → 2 transition is ΔL = (2 - 5)ћ = -3ћ in magnitude 3ћ. In this simple Bohr view the photon must carry away the angular momentum difference (magnitude 3ћ). (Note: this is a semiclassical Bohr estimate; full quantum theory imposes selection rules that limit allowed Δℓ values.)

LONG ANSWER TYPE QUESTIONS

Q.24. The first four spectral lines in the Lyman serics of a H-atom are λ = 1218 Å, 1028 Å, 974.3 Å and 951.4 Å. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.
Ans: Taking nuclear motion into account, the spectral frequencies scale with the reduced mass μ. Wavelength λ ∝ 1/μ. For hydrogen (mass M_H ≈ M) and deuterium (mass M_D ≈ 2M) the reduced masses give the ratio
λ_D/λ_H = μ_H/μ_D ≈ 0.9997277 (using μ ≈ m_e(1 - m_e/M)).
Thus λ_D ≈ 0.9997277 × λ_H and the shift Δλ = λ_D - λ_H ≈ -0.0002723 × λ_H.
Applying this to the four lines:
λ_D1 ≈ 0.9997277 × 1218 Å ≈ 1217.67 Å (shift ≈ -0.33 Å).
λ_D2 ≈ 0.9997277 × 1028 Å ≈ 1027.72 Å (shift ≈ -0.28 Å).
λ_D3 ≈ 0.9997277 × 974.3 Å ≈ 974.03 Å (shift ≈ -0.27 Å).
λ_D4 ≈ 0.9997277 × 951.4 Å ≈ 951.14 Å (shift ≈ -0.26 Å).
These small reductions reflect the slightly larger reduced mass of deuterium. The detailed reduced-mass steps are shown in the image placeholders above.

Q.25. Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in ¹H and ²H. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass µ, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here µ = m_eM /(m_e +M) where M is the nuclear mass and m_e is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in ¹H and ²H. (Mass of ¹H nucleus is 1.6725 × 10^-27 kg, Mass of ²H nucleus is 3.3374 × 10^-27 kg, Mass of electron = 9.109 × 10^-31 kg.)
Ans: The wavelength scales inversely with the reduced mass: λ ∝ 1/μ. Therefore the fractional change in wavelength is Δλ/λ ≈ (μ_H/μ_D - 1). Using μ ≈ m_e(1 - m_e/M) for M ≫ m_e, we find numerically:
m_e/M_H ≈ 5.446 × 10^-4, m_e/M_D ≈ 2.730 × 10^-4.
Thus μ_H ≈ m_e(0.9994554), μ_D ≈ m_e(0.9997270).
So μ_H/μ_D ≈ 0.9997284 and Δλ/λ = 0.0002716 ≈ 2.716 × 10^-4 = 0.02716%.
Therefore the percentage difference in the wavelength of the first Lyman line between ¹H and ²H is about 0.027%. The image placeholders show the algebraic steps leading to this result.

Q.26. If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1 Å, and (ii) R = 10 Å.
Ans: If the proton radius R is much smaller than the Bohr radius a₀ (a₀ ≈ 0.53 Å), the electron effectively sees the nucleus as a point charge and the usual Bohr ground-state energy E₁ ≈ -13.6 eV holds to good approximation. Thus for (i) R = 0.1 Å (< />₀) the ground-state energy remains essentially -13.6 eV.
For (ii) R = 10 Å (≫ a₀) the nuclear charge is spread over a volume much larger than the normal electron orbit. Inside a uniformly charged sphere the electrostatic potential is less deep and varies less steeply with r; the electron is therefore far less tightly bound. Using the potential of a uniformly charged sphere (shown in the image placeholders) one finds a much smaller binding energy (a few eV in magnitude) rather than -13.6 eV. The detailed algebra and intermediate expressions are retained in the image placeholders above; the qualitative conclusion is: for R ≪ a₀ the usual -13.6 eV holds, while for R ≫ a₀ the binding energy is greatly reduced (smaller magnitude).

Q.27. In the Auger process an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an n = 4 Auger electron emitted by Chromium by absorbing the energy from a n = 2 to n = 1 transition.
Ans: The energy released in the 2 → 1 transition is E₁ - E₂ = (3/4) Z² R (in energy units where R is 13.6 eV for hydrogen). This energy is transferred to the electron in n = 4 which requires an amount equal to its binding energy |E₄| = Z² R /4² = (1/16) Z² R to remove it. The kinetic energy of the ejected (Auger) electron is the difference: K.E. = (E₁ - E₂) - |E₄|, which after substitution and using Z = 24 and R = 13.6 eV gives the numerical result shown in the image placeholders. The images contain the stepwise substitution and numeric evaluation leading to the K.E. value (~5.39 × 10³ eV shown).

Q.28. The inverse square law in electrostatics is 

 for the force between an electron and a proton. The (1/r) dependence of |F| can be understood in quantum theory as being due to the fact that the 'particle' of light (photon) is massless. If photons had a mass m_p, force would be modified to 

 where 

 Estimate the change in the ground state energy of a H-atom if m_p were 10^-6 times the mass of an electron.
Ans: If the photon had a small mass the potential would become Yukawa-like and slightly deviate from 1/r at large distances. For m_p = 10^-6 m_e the photon Compton wavelength λ = h/(m_pc) is still much larger than the Bohr radius, so the correction to the Coulomb potential is extremely small at atomic scales. Carrying out the perturbative estimate (the algebra is shown in the image placeholders) gives a tiny fractional change in the ground state energy of order 27.2 λ r_A eV (as obtained in the image steps), which is negligible compared to 13.6 eV for the chosen m_p. The detailed derivation and numerical estimate are retained in the placeholders above.

Q.29. The Bohr model for the H-atom relies on the Coulomb's law of electrostatics. Coulomb's law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb's law between two opposite charge + q₁, -q₂ is modified to

Calculate in such a case, the ground state energy of a H-atom, if ε = 0.1, R₀ = 1Å.
Ans: For a short-range modification of Coulomb's law the ground-state radius and energy change. The image placeholders contain the detailed algebra using the modified force law inside r ≤ R₀ and matching to the usual form outside. Substituting ε = 0.1 and R₀ = 1 Å into those expressions and following Bohr's quantisation yields the modified ground-state radius r₁, the velocity and the kinetic and potential energies; the numerical evaluation in the placeholders gives a total energy approximately -11.4 eV for the chosen parameters. The full derivation with intermediate numerical steps is preserved in the image placeholders above.