Atoms Exemplar Solutions Class 12 - Free PDF

Correct option: (c) 18 pm.

Concept used. In Bohr's model the radius of the nth permitted orbit in a hydrogen-like atom (one electron, nuclear charge +Ze) is r_n = n² a₀Z, where a₀ = 53 pm is the Bohr radius (the ground-state radius of ordinary hydrogen, Z=1, n=1). For lithium the atomic number is Z=3, so Li⁺⁺ (a lithium atom that has lost two of its three electrons) is a one-electron hydrogen-like ion with Z=3.

1. Identify the parameters for Li⁺⁺ in the ground state: Z = 3, n = 1.
2. Substitute into the Bohr radius formula: r₁ = 1² × a₀3 = a₀3.
3. Plug in a₀ = 53 pm: r₁ = 53 pm3 = 17.67 pm ≈ 18 pm.
4. Compare with the options: (a) 53 matches hydrogen, not Li⁺⁺; (b) 27 would correspond to Z=2 (He⁺); (d) 13 is too small for Z=3.

Option (c): r ≈ 18 pm.

Correct option: (c) the frame in which the electron is at rest is not inertial.

Concept used. Newton's laws (and the derivation of Bohr's orbits, which uses centripetal force = Coulomb force) only hold in an inertial frame: a frame moving at constant velocity relative to the fixed stars, with no proper acceleration. In a hydrogen atom the electron orbits the (nearly fixed) proton; in the electron's rest frame the proton accelerates around the electron, so that frame itself is non-inertial.

1. Recall the Bohr derivation. The radius formula r_n = n² a₀/Z and energy E_n ∝ m arise from balancing mv²r = 14π₀e²r² and quantising L = mvr = n. Both steps assume Newton's second law holds, i.e. the lab frame is inertial.
2. In the proton's frame the electron moves in a closed orbit, so the proton frame is (nearly) inertial; fine.
3. Switch to the electron's rest frame. Here the electron itself is following a circle in the lab frame, so its rest frame is constantly accelerating. Newton's laws need fictitious-force corrections; the naive replacement m → M in the binding energy ignores those, so the formula fails.
4. Eliminate the other options: (a) n is still integer (the quantisation rule does not depend on which particle is moving); (b) Bohr-quantisation applies to any orbiting particle in an inertial frame; (d) approximate circular orbits do exist relative to the centre of mass.

Option (c): the electron's rest frame is non-inertial, so the simple Bohr derivation cannot be transplanted into it.

Correct option: (a) of the electrons not being subject to a central force.

Concept used. A central force is one that points always toward (or away from) a fixed centre and whose magnitude depends only on the distance from that centre. Bohr's derivation (circular orbit + quantisation) is built entirely on the assumption that the electron experiences exactly such a force, supplied by the nucleus alone.

1. For hydrogen, the electron feels only the Coulomb attraction from the proton, which is purely radial; a central force. So Bohr's model works.
2. In a many-electron atom, each electron feels the nucleus plus the repulsions from all the other electrons. Those electron-electron forces do not point toward the nucleus and depend on the angular positions of all the other electrons. The net force is no longer central.
3. Without a central force, angular momentum about the nucleus is not conserved and circular Bohr orbits with fixed radii cease to be exact solutions.
4. Why the other options are weaker: (b) collisions are not the primary reason; (c) screening is a consequence of, not the underlying cause of, the central-force breakdown; (d) the electron-nucleus force is still Coulombic, it is the electron-electron pieces that spoil centrality.

Option (a): in a many-electron atom the net force on each electron is not central.

Correct option: (a) because the Bohr model gives incorrect values of angular momentum.

Concept used. The Bohr model postulates L = n, so the ground-state (n=1) angular momentum is . But quantum mechanics (the more accurate theory) gives L = √(+1) , where = 0, 1, , n-1. For n=1 the only allowed is = 0, giving L = 0, not .

1. State the Bohr ground-state value: L_Bohr = 1·= .
2. State the correct quantum-mechanical value for n=1, =0: L = √0(0+1) = 0.
3. Since L=0 in the true (Schrödinger) treatment, there is no non-zero vector to point in different directions; the infinite-orbit paradox disappears. So the premise of the question is wrong: the Bohr value is incorrect for the ground state.
4. Reject the other options: (b) is true for any system but does not address the orientation issue; (c) confuses orbital angular momentum with spin; (d) is geometrically meaningless (no "horizontal" in an isolated atom).

Option (a): the Bohr model's ground-state value L= is itself wrong; the correct quantum-mechanical ground-state value is L=0.

Correct option: (a) is not important because nuclear forces are short-ranged.

Concept used. The strong nuclear force that binds protons and neutrons inside a nucleus has an effective range of about 1 fm = 10^-15 m. Beyond a few fm it drops essentially to zero. In contrast, atomic separations in a diatomic molecule are of the order of 1 = 10^-10 m, which is 10⁵ times larger than the nuclear-force range.

1. Estimate the inter-nuclear distance in O₂: bond length ≈ 1.2 = 1.2 × 10^-10 m.
2. Compare with the range of the strong nuclear force: r_nuc ∼ 1 fm = 1 × 10^-15 m.
3. Ratio: r_bondr_nuc = 1.2 × 10^-101 × 10^-15 = 1.2 × 10⁵. The nuclei are ∼ 10⁵ nuclear-force ranges apart, so the strong force between them is utterly negligible.
4. Eliminate the others: (b) and (c) are factually false at molecular distances; (d) is irrelevant because what matters is the nuclei being far apart, not their internal composition.

Option (a): nuclear forces are short-ranged (∼ 1 fm) while O₂ nuclei are ∼ 1 apart.

Correct option: (a) 10.20 eV.

Concept used. In an inelastic collision, kinetic energy can be converted into internal energy. For two hydrogen atoms, the minimum internal excitation that is allowed by quantum mechanics is the transition n=1 → n=2, which costs Δ E = E₂ - E₁ = -13.62² - (-13.61²) = -3.4 + 13.6 = 10.2 eV. Energies less than 10.2 eV cannot be absorbed (the spectrum is discrete); transitions to n=3 cost 12.09 eV, ionisation 13.6 eV. So if the collision delivers only just enough energy to excite one atom, the smallest non-zero loss in KE is 10.2 eV.

1. Compute the n=1 → n=2 excitation energy: Δ E₁₂ = 13.6(1 - 1/4) = 13.6 × 3/4 = 10.2 eV.
2. The question asks for the maximum reduction in combined KE consistent with "inelastic" with the smallest available internal channel; interpreted as the smallest single-atom excitation. (Options (c) 13.6 corresponds to full ionisation; (d) 27.2 to ionising both atoms; (b) 20.4 = 2 × 10.2 to exciting both atoms, which requires both atoms to absorb exactly 10.2; possible only at very specific kinematics.)
3. The Exemplar's answer-key reading: in a typical inelastic collision the most that one atom can take and still leave the atoms intact in low-lying states is one n=1 → n=2 excitation, Δ KE = 10.2 eV.

Option (a): Δ KE_max = 10.20 eV (energy to excite one H atom from n=1 to n=2).

Correct option: (a) in general to any of the states with lower energy.

Concept used. An excited state of an atom is a state with energy above the ground state. Excited states are unstable; they decay spontaneously by emitting a photon, with a photon frequency ν set by hν = E_initial - E_final, where E_final < E_initial. The transition can be to any lower-energy state allowed by quantum selection rules; not only to the ground state.

1. Start from an atom in excited state |i with energy E_i. There may be several lower-energy states |f with E_f < E_i.
2. Each decay channel has its own probability per unit time (Einstein A coefficient). A given atom selects one channel randomly; an ensemble distributes across all of them.
3. Decay is spontaneous, so no external field (option b) or collision (option d) is required. It does not happen simultaneously for all atoms (option c); each atom has its own random decay time.
4. Hence "in general to any of the states with lower energy" is the right description, giving rise to the multi-line emission spectrum we observe.

Option (a).

Correct options: (a) and (c).

Concept used. The hydrogen molecular ion H₂⁺ has two centres of positive charge (the two protons) and a single electron. The potential the electron sees is the sum of two Coulomb wells, one at each proton: V(r) = -e²4π₀ |r - r_A| -e²4π₀ |r - r_B|. This potential is not spherically symmetric, so simple Bohr-style circular orbits do not exist.

1. No spherical symmetry ⇒ angular momentum about any single point is not conserved. The electron's motion is complicated; it does not trace a circle. So (a) is correct.
2. The lowest-energy electronic state has a wavefunction that wraps around both protons (the bonding _g orbital). The electron is shared between the two centres. So (c) is correct.
3. (b) is wrong. A naive Z=2 scaling would give energy 2² = 4 times hydrogen, not 2⁴ = 16. Moreover the two-centre problem does not reduce to a single-Z formula at all.
4. (d) is wrong. H₂⁺ is a stable bound species (it is observed in mass spectrometers); it does not spontaneously decay into p + H.

Correct: (a), (c).

Correct options: (a) and (b).

Concept used. Conservation of energy must hold in any scattering process. The free electron + free proton system has non-negative total energy (E_i = KE_rel ≥ 0); a bound H-atom has negative total energy (E_f ≤ -13.6 eV). The positive energy difference E_i - E_f ≥ 13.6 eV cannot disappear, so capture is impossible unless that energy is simultaneously carried away by an emitted photon.

1. Energy balance. Initial: E_i = KE_rel ≥ 0. Final: E_f = -13.6/n² eV ≤ -13.6 eV for n=1. The deficit E_i - E_f ≥ 13.6 eV cannot vanish; energy conservation alone forbids the bare two-body capture → (a) is correct.
2. Equivalently, the only way to balance the books is for the excess energy to be radiated as a photon at the moment of capture. Without that simultaneous emission of radiation, capture is impossible → (b) is correct.
3. (c) Linear momentum can be conserved by recoil of the H-atom (or by the emitted photon), so momentum conservation is not the obstruction.
4. (d) Angular momentum can also be balanced via the photon's spin and orbital angular momentum; in any case, it is not the headline reason the bare e + p → H capture fails.

Correct: (a), (b).

Correct options: (a) and (b).

Concept used. Bohr's model is built for a single electron orbiting a single nucleus under a central Coulomb force. Any deviation from that setup invalidates the simple model.

1. Molecular hydrogen H₂ has two nuclei. The single-centre Bohr picture cannot describe it → (a) is correct.
2. Helium (Z=2) has two electrons. The repulsion between the electrons (and the loss of centrality of the net force on each) means simple Bohr is not directly applicable to neutral He → (b) is correct.
3. Temperature does not enter the Bohr derivation at all; the atom is treated in isolation. So (c) is wrong.
4. Bohr's spectrum is purely discrete (one line per n_i → n_f transition); no continuum. So (d) is wrong.

Correct: (a), (b).

Correct options: (b) and (d).

Concept used. The Balmer series is defined as the set of spectral lines emitted by hydrogen when an electron drops from any level n ≥ 3 to n=2 (the first excited state). The Rydberg formula gives the wavelengths as 1λ = R (12² - 1n²), n = 3, 4, 5,

1. (a) is wrong because falling to the ground state (n=1) produces the Lyman series, not Balmer.
2. (b) is correct by definition: transitions to the first excited state n=2 give the Balmer series.
3. (c) is wrong: only transitions ending at n=2 count as Balmer.
4. For (d), look at 1/λ = R(1/4 - 1/n²). As n → ∞, 1/λ → R/4; the series limit. The line spacings shrink rapidly: aligned n=3: 1/λ &= R(1/4 - 1/9) = 5R/36,
   n=4: 1/λ &= R(1/4 - 1/16) = 3R/16,
   n=5: 1/λ &= R(1/4 - 1/25) = 21R/100,
   n=∞: 1/λ &= R/4. aligned Successive 1/λ values approach R/4 ever more closely, so the high-frequency end of the series is densely packed. Hence (d) is correct.

Correct: (b), (d).

Correct options: (b) and (d).

Concept used. Resonant absorption occurs when the incoming photon's energy exactly matches the gap between two atomic levels: hν = E_f - E_i. Here the photon energy is precisely E₂ - E₁, which matches the n=1 → n=2 transition. The absorption is real but its rate is finite, so not every atom is excited at once.

1. Match the photon energy to the available transitions. hν = E₂ - E₁ matches the 1 → 2 transition exactly, but is too small for the 1 → 3 transition (which needs E₃ - E₁ = 12.09 eV > 10.2 eV).
2. Therefore some atoms absorb the photon and jump to n=2 → (b) is correct.
3. No atoms can absorb a single photon and jump to n=3, since the photon energy is insufficient → (d) is correct.
4. (a) is wrong: photons are absorbed.
5. (c) is too strong: absorption is a probabilistic, finite-rate process. At any finite intensity and duration, only a fraction of atoms is excited; and excited atoms also spontaneously emit back to n=1, setting up a dynamic equilibrium where not all atoms are in n=2.

Correct: (b), (d).

Correct options: (c) and (d).

Concept used. Bohr's model assumes a single electron in a purely central Coulomb potential created by the nucleus. Neutral He has two electrons, so each electron sees not just the nucleus but also the other electron's repulsion; and that repulsion has no fixed centre.

1. He has Z=2 and two electrons. Compared to H, it has "one more electron" → (c) is correct.
2. Because the second electron is not at the nucleus, the electron-electron Coulomb force on a given electron is not radial; the net force on each electron is no longer central → (d) is correct.
3. (a) "Inert" is a chemical property and has no bearing on the validity of Bohr's mechanical model.
4. (b) Neutrons are electrically neutral and inside the nucleus they do not affect the electrostatic force seen by electrons outside.

Correct: (c), (d).

Concept used. The mass-energy equivalence of Einstein, E = mc², states that any bound system has less mass than the sum of the masses of its free constituents, by exactly m_def = B/c², where B is the binding energy (the energy required to dissociate the system into free constituents).

1. For hydrogen in the ground state, the binding energy is B = 13.6 eV.
2. Convert to mass deficit: m_def = Bc² = 13.6 × 1.6 × 10^-19 J (3 × 10⁸)² m²/s² = 2.176 × 10^-189 × 10¹⁶ = 2.42 × 10^-35 kg.
3. Therefore m_H-atom = m_p + m_e - 2.42 × 10^-35 kg , about 10^-5 of the electron mass; very small but present. The "missing" mass was carried away as a photon when the proton and electron combined into the bound H-atom.

The bound H-atom has a mass deficit of m_def = B/c² ≈ 2.42 × 10^-35 kg, equal to the binding energy 13.6 eV divided by c².

Concept used. For a hydrogen-like ion (one electron, nucleus of charge +Ze and mass M) the Bohr-model energy is E_n = -μ Z² e⁴8₀² n² h², where μ = m_e M/(m_e + M) is the reduced mass. The dependence on M enters only through μ and is very weak when M ≫ m_e.

1. Both He⁴ and He³ are helium isotopes with Z=2; removing one electron leaves He⁺, a hydrogen-like ion with Z=2. The only difference between the two cases is the nuclear mass: M(He⁴) ≈ 4 u, M(He³) ≈ 3 u.
2. Reduced masses: ₄ = m_e M₄m_e + M₄ ≈ m_e(1 - m_eM₄), ₃ ≈ m_e(1 - m_eM₃). With m_e/M ∼ 1/7000, the corrections are ∼ 10^-4.
3. Relative difference: ₄ - ₃m_e ≈ m_eM₃ - m_eM₄ ≈ 13· 1836·14 · 1(order 1) ∼ 10^-5. Levels shift by about 1 part in 10⁵, far smaller than ordinary spectral resolution.

Because the Bohr energies depend on nuclear mass only through the reduced mass μ; for both He isotopes M ≫ m_e, so μ ≈ m_e to within ∼ 10^-4. The energy levels therefore differ negligibly.

Concept used. An atomic electron is a charged particle. The transition between levels involves an oscillation of charge, which couples to the electromagnetic field. Conservation of energy and momentum, plus the absence of any other coupling, force the emitted energy to take the form of a photon.

1. The electron carries charge -e. Its only first-order coupling to the surrounding vacuum is electromagnetic (the weak and strong interactions are much weaker for atomic-scale energies, and gravity is negligible).
2. During the transition, the electron's wavefunction changes in a way that produces a time-varying dipole moment. A time-varying charge distribution radiates electromagnetic waves; that is classical electrodynamics, refined by QED to photon emission.
3. Energy and (linear and angular) momentum are conserved by the emitted photon: hν = E_i - E_f, p_γ balanced by atomic recoil, photon spin matches the Δ selection rule.
4. No other channel is available: the atom is isolated, so no kinetic energy can be given to surroundings; no heat bath is present; only the radiation field is available. Hence the energy is emitted as a photon.

Because the electron's only relevant coupling to the surroundings is electromagnetic, the transition energy must be emitted as a photon; there is no other channel available in an isolated atom.

Concept used. The Bohr formula for hydrogen-like energies depends on the charges only through the product q_e q_p that appears in Coulomb's law, and ultimately through the combination (q_e q_p)² that appears in the energy: E_n = -m_e (q_e q_p)²8₀² n² h². Therefore only the product q_e q_p matters, not the individual charges.

1. In standard hydrogen, q_p = +e, q_e = -e, so |q_e q_p| = e².
2. In the modified hydrogen of this question: q_p = (+4/3)e, q_e = (-3/4)e, so |q_e q_p| = 43e × 34e = e². The product is unchanged.
3. Substitute into the Bohr energy: since (q_e q_p)² = e⁴ in both cases, E_n is the same.
4. Therefore the Bohr formula (in terms of e) remains unchanged.

Yes, the Bohr formula remains unchanged, because |q_e q_p|=(4/3)(3/4)e² = e² exactly matches the standard case, so the Coulomb-derived energies are identical.

Concept used. In Bohr's model, the angular momentum and energy of the nth level are both fixed by the single quantum number n: L_n = n, E_n = -13.6n² eV. So L and E are one-to-one through n.

1. Two atoms with different energies must be in different n: E_n ≠ E_n' ⇒ n ≠ n'.
2. In Bohr, different n also means different L: L_n = n≠ n'= L_n'.
3. Therefore different energies imply different angular momenta in the Bohr model. It is not possible to have different energies but the same L.
4. (Note: in the full quantum theory, the answer is different because L is set by and E is set by n, which are independent; but the question asks about the Bohr model.)

No. In the Bohr model both E and L are fixed by n, so different energies imply different angular momenta.

Concept used. For a two-body Coulomb system, the Bohr-model energy is E_n = -μ e⁴8₀² n² h² = -μm_e· 13.6n² eV, where μ = m₁ m₂/(m₁ + m₂) is the reduced mass. For ordinary hydrogen, m₂ = m_p ≫ m_e, so μ ≈ m_e. For positronium, the two particles have equal mass m_e.

1. Compute the reduced mass of positronium: μ = m_e · m_em_e + m_e = m_e²2 m_e = m_e2.
2. Compute the ratio μ/m_e: μm_e = 12.
3. Substitute in the Bohr formula for n=1: E₁^Ps = -μm_e· 13.6 eV = -12· 13.6 eV = -6.8 eV.
4. Sanity check: the positronium "Bohr radius" is a₀^Ps = (m_e/μ) a₀ = 2 a₀ ≈ 106 pm, twice ordinary hydrogen; also consistent.

E₁^Positronium = -6.8 eV (half the hydrogen ground-state energy).

Concept used. Without electron-electron repulsion, the two electrons of helium are independent. Each electron moves in the central Coulomb field of the Z=2 nucleus, so each is governed by the hydrogen-like Bohr formula with Z=2. The total ground-state energy is then twice the single-electron ground-state energy.

1. Single-electron energy in the field of the He nucleus (Z=2, n=1): E₁(Z=2) = -13.6 × Z²n² eV = -13.6 × 4 eV = -54.4 eV.
2. Both electrons sit in this n=1 level (allowed because they are non-interacting; in real He they would also fit, since electrons are fermions and the two have opposite spins).
3. Total ground-state energy: E_{He, ground} = 2 × E₁(Z=2) = 2 × (-54.4) eV = -108.8 eV.
4. Sanity check: the experimental He ground-state energy is -79.0 eV; the difference -108.8 - (-79.0) = -29.8 eV is the true electron-electron repulsion energy, which we set to zero in this idealisation.

E_{He, no repulsion} = -108.8 eV.

Concept used. An electron moving in a circular Bohr orbit is equivalent to a tiny current loop. The current carried by a single charge -e orbiting with period T is I = eT = eν = eω2π, where ν = 1/T is the orbital frequency and ω the angular frequency.

1. For the ground state of H (n=1), the orbital radius is the Bohr radius: r = a₀ = 5.29 × 10^-11 m.
2. The orbital speed comes from Bohr's quantisation m_e v r = n with n=1: v = m_e a₀. Substitute = 1.055 × 10^-34 J s, m_e = 9.11 × 10^-31 kg, a₀ = 5.29× 10^-11 m: v = 1.055 × 10^-349.11 × 10^-31 × 5.29 × 10^-11 = 1.055 × 10^-344.819 × 10^-41 = 2.19 × 10⁶ m/s.
3. Period of one revolution: T = 2π rv = 2π × 5.29 × 10^-112.19 × 10⁶ = 3.324 × 10^-102.19 × 10⁶ = 1.518 × 10^-16 s.
4. Current: I = eT = 1.6 × 10^-191.518 × 10^-16 = 1.054 × 10^-3 A ≈ 1.05 mA.

I ≈ 1.05 × 10^-3 A ≈ 1.05 mA.

Concept used. The Rydberg formula for transitions n+p → n gives the emission frequency _p = c R (1n² - 1(n+p)²), p = 1, 2, 3, For large n we Taylor-expand 1/(n+p)² in powers of p/n to extract the dominant behaviour.

1. Expand 1(n+p)² for p ≪ n: 1(n+p)² = 1n²(1 + pn)^-2 = 1n²(1 - 2pn + 3p²n² - ).
2. Substitute into the Rydberg formula: _p = cR[1n² - 1n²(1 - 2pn + 3p²n² - )].
3. Simplify by cancelling 1/n²: _p = cR[2pn³ - 3p²n⁴ + ].
4. Keep the leading term (valid when n ≫ p): _p ≈ 2cR pn³. Hence ₁1 = ₂2 = ₃3 = = 2cRn³, showing _p ∝ p, i.e. the first few frequencies are in the ratio 1:2:3:; they are approximate harmonics.

For n ≫ 1, _p ≈ (2cR/n³) p, so successive emission frequencies ₁:₂:₃ = 1:2:3, i.e. approximate harmonics.

Concept used. The H_γ line is the third line of the Balmer series, corresponding to the transition n=5 → n=2 (Balmer lines: H_α is 3→ 2, H_β is 4 → 2, H_γ is 5 → 2). For a ground-state atom to emit H_γ on its way down, it must first be excited to n=5. The minimum energy required is the excitation energy E₅ - E₁. The photon emitted in the 5 → 2 transition then carries angular momentum equal to the difference of the atomic angular momenta (in the Bohr picture L_n = n), or, equivalently, the photon itself carries of intrinsic spin.

1. Bohr energies (in eV): E_n = -13.6/n². E₁ = -13.6, E₅ = -13.6/25 = -0.544.
2. Minimum energy to raise the atom to n=5: Δ E = E₅ - E₁ = -0.544 - (-13.6) = 13.056 eV ≈ 13.06 eV.
3. Angular momentum of the photon. In the Bohr model the atom's orbital L changes from L₅ = 5 to L₂ = 2 in the emission, so the photon must carry Δ L_atom = L₅ - L₂ = 5- 2= 3. (In the full quantum theory each photon carries spin , and a single 5 → 2 transition is allowed only when Δ= ± 1. The Bohr-style answer expected by the Exemplar is 3.)

Minimum excitation energy ≈ 13.06 eV; angular momentum of H_γ photon = 3 (Bohr-model estimate).

Concept used. The Rydberg constant for an atom is R = R_∞ · μm_e, where μ = m_e M/(m_e + M) and M is the nuclear mass. The wavelength of a Rydberg transition is λ ∝ 1/R, so _DH = R_HR_D = _HD.

1. Reduced masses, using m_e = 9.109 × 10^-31 kg, M_H = 1.6725 × 10^-27 kg, M_D = 3.3374 × 10^-27 kg: _H = m_e M_Hm_e + M_H = (9.109× 10^-31)(1.6725× 10^-27) (9.109× 10^-31) + (1.6725× 10^-27). Denominator ≈ 1.67341 × 10^-27 kg. Numerator ≈ 1.5234 × 10^-57. So _H ≈ 9.104 × 10^-31 kg.
2. Similarly for deuterium: _D = (9.109× 10^-31)(3.3374× 10^-27) (9.109× 10^-31) + (3.3374× 10^-27) ≈ 9.106× 10^-31 kg.
3. Ratio: _HD = 9.1049.106 ≈ 0.99973. Equivalently, _D = 0.99973 × _H, i.e. the deuterium lines are shorter by about 0.0273%.
4. Shift for each line λ = _H - _D = _H (1 - 0.99973) = 0.00027 _H: aligned _H = 1218 : && λ &≈ 0.00027 × 1218 ≈ 0.33 ,
   _H = 1028 : && λ &≈ 0.00027 × 1028 ≈ 0.28 ,
   _H = 974.3 : && λ &≈ 0.00027 × 974.3 ≈ 0.26 ,
   _H = 951.4 : && λ &≈ 0.00027 × 951.4 ≈ 0.26 . aligned
5. Direction: deuterium lines are slightly shorter than hydrogen lines (blueshift), because _D > _H gives a slightly larger Rydberg constant.

Wavelength shifts λ ≈ 0.33, 0.28, 0.26, 0.26 for the first four Lyman lines (deuterium lines blueshifted relative to hydrogen).

Concept used. The Rydberg-formula wavelength scales inversely with the reduced mass: 1λ = R_∞ · μm_e(1n₁² - 1n₂²). For the same transition (same n₁, n₂), _HD = _DH, so the percentage difference (_H - _D)/_H × 100% is the same for every line.

1. Compute _H. With m_e = 9.109× 10^-31, M_H = 1.6725× 10^-27: _H = m_e M_Hm_e + M_H = (9.109× 10^-31)(1.6725× 10^-27) 9.109× 10^-31 + 1.6725× 10^-27. Denominator = 1.67341 × 10^-27 kg. Numerator = 1.5234 × 10^-57. _H = 1.5234× 10^-571.67341× 10^-27 = 9.1041 × 10^-31 kg.
2. Compute _D: _D = (9.109× 10^-31)(3.3374× 10^-27) 9.109× 10^-31 + 3.3374× 10^-27 = 9.1065 × 10^-31 kg. (Denominator = 3.33831 × 10^-27.)
3. Ratio: _DH = 9.10659.1041 = 1.000264. Therefore _H/_D = 1.000264, i.e. the deuterium line is shorter by a factor 1/1.000264 = 0.999736.
4. Percentage difference: _H - _DH = 1 - 11.000264 = 1 - 0.999736 = 2.64 × 10^-4 = 0.0264%. Some textbooks round to 0.0273%; the small variation is in the round-off of μ.

Percentage difference in wavelength ≈ 0.026% (deuterium line slightly shorter than hydrogen line).

Concept used. Inside a uniformly charged sphere of radius R and total charge +e, Gauss's law gives the field as growing linearly with distance from the centre: E(r) = e4π₀· rR³ (r < R), while outside (r > R) the field is the usual Coulomb field of a point charge.

The corresponding potential energy of an electron at distance r is U(r) = cases e²8π₀ R(r²R² - 3), & r < R,
[6pt] -e²4π₀ r, & r ≥ R, cases (reference: U(∞) = 0, U continuous at r=R).

We compare the Bohr ground-state radius a₀ = 0.53 with each R. If a₀ > R, the electron orbits outside the proton and the usual Bohr formula applies. If a₀ < R, the electron orbits inside the proton and a new derivation is needed.

Case (i): R = 0.1 < a₀ = 0.53

1. Since the Bohr radius a₀ = 0.53 is much larger than R = 0.1 , the electron in the ground state lies outside the charge distribution. The proton looks point-like to the electron.
2. Therefore the standard Bohr formula applies: E₁ = -13.6 eV.
3. There is essentially no change from the point-proton case because the electron never penetrates the charge sphere.

Case (ii): R = 10 ≫ a₀

Now the electron orbits well inside the proton's charge distribution. We redo the Bohr derivation using the interior potential.

1. Inside the sphere (r < R), the field on the electron is E(r) = (e/4π₀) r / R³. The Coulomb force on the electron has magnitude F(r) = e²4π₀· rR³. This is a linear-restoring (Hooke-like) force, just like a 3D harmonic oscillator.
2. Circular-orbit condition (centripetal = Coulomb): m_e v²r = e² r4π₀ R³ ⇒ v² = e² r²4π₀ m_e R³.
3. Bohr quantisation m_e v r = n with n=1: v = m_e r. Equate with v² from step 2: ²m_e² r² = e² r²4π₀ m_e R³.
4. Solve for r⁴: r⁴ = 4π₀ ² R³m_e e² = a₀ R³, using a₀ = 4π₀ ² /(m_e e²). So r = (a₀ R³)^1/4. Plug in a₀ = 0.53 , R = 10 : r = (0.53 × 10³)^1/4 = (530)^1/4 ≈ 4.80 . Since r = 4.80 < R = 10 , the assumption "r < R" is self-consistent.
5. Compute the orbital kinetic energy: KE = 12 m_e v² = ²2 m_e r². Numerically, with r = 4.80× 10^-10 m: aligned KE &= (1.055× 10^-34)² 2 × 9.11× 10^-31 × (4.80× 10^-10)²
   &= 1.113× 10^-684.199× 10^-49 = 2.65 × 10^-20 J. aligned Convert: 2.65 × 10^-20/1.6× 10^-19 = 0.166 eV.
6. Compute the potential energy inside the sphere at r = 4.80 : U = e²8π₀ R(r²R² - 3). Numerator prefactor: aligned e²8π₀ R &= 12· e²4π₀ R
   &= 12· (1.6× 10^-19)² × 9× 10⁹ 10× 10^-10
   &= 12· 2.304 × 10^-19
   &= 1.152 × 10^-19 J = 0.72 eV. aligned Bracket: r²/R² - 3 = (4.80/10)² - 3 = 0.2304 - 3 = -2.770. So U = 0.72 × (-2.770) = -1.99 eV.
7. Total energy: E = KE + U = 0.166 - 1.99 = -1.83 eV. (Bohr ground-state energy is dramatically less negative; the atom is much less bound; because the linear restoring force is weaker than the 1/r² Coulomb force at these distances.)

(i) For R=0.1 : standard Bohr value, E₁ = -13.6 eV. (ii) For R=10 : r ≈ 4.80 , E₁ ≈ -1.83 eV.

Concept used. In the Auger process an inner-shell vacancy is filled by an electron from a higher level, and the energy released is given to another bound electron (the Auger electron), which is then ejected. Energy conservation: the energy released in the inner transition equals the binding energy of the Auger electron plus its kinetic energy after ejection.

For chromium Z = 24. In the simplest hydrogen-like (Bohr-model) approximation, energy levels scale as E_n = -13.6 Z²n² eV.

1. Compute the energy released in the n=2 → n=1 transition of Cr in the hydrogen-like approximation: Δ E₂₁ = E₁ - E₂ = -13.6 Z² (1 - 14) = 13.6 × 24² × 34 eV. Compute step by step: 24² = 576. 13.6 × 576 = 7833.6. 7833.6 × 3/4 = 5875.2 eV. So Δ E₂₁ = 5875.2 eV.
2. This energy is delivered to the n=4 Auger electron. The binding energy of an n=4 electron in the Cr field (still hydrogen-like approximation) is |E₄| = 13.6 Z²n² = 13.6 × 57616 = 7833.616 = 489.6 eV.
3. Energy conservation: the released energy goes partly into unbinding the n=4 electron and partly into its kinetic energy: Δ E₂₁ = |E₄| + KE ⇒ KE = Δ E₂₁ - |E₄| = 5875.2 - 489.6 = 5385.6 eV.

KE_Auger = Δ E₂₁ - |E₄| = 5875.2 - 489.6 ≈ 5385.6 eV ≈ 5.39 keV.

Concept used. A massive-photon (Yukawa-type) modification of Coulomb's law replaces 1/r by an exponentially screened potential U(r) = -e²4π₀· e^{-λ r}r, λ = m_p c. For λ r ≪ 1 the modification is a tiny perturbation; we can estimate the change in the ground-state energy by computing Δ U on the unperturbed Bohr wavefunction (equivalently, evaluating Δ U at the Bohr radius).

1. Compute λ for m_p = 10^-6 m_e: λ = (10^-6)(9.11× 10^-31)(3× 10⁸) 1.055× 10^-34. Numerator = 10^-6 × 9.11× 10^-31 × 3× 10⁸ = 10^-6 × 2.733× 10^-22 = 2.733 × 10^-28 kg m/s. Therefore λ = 2.733× 10^-281.055× 10^-34 = 2.591 × 10⁶ m^-1. i.e. λ ≈ 2.6 × 10⁶ m^-1, with 1/λ ≈ 3.86× 10^-7 m.
2. Compare λ r at the Bohr radius r = a₀ = 5.29× 10^-11 m: λ a₀ = 2.591× 10⁶ × 5.29× 10^-11 = 1.37 × 10^-4. Very small: the exponential e^{-λ r} ≈ 1 - λ r.
3. Expand the modified potential to first order in λ r: U(r) = -e²4π₀·e^{-λ r}r ≈ -e²4π₀·1 - λ rr = -e²4π₀ r + e² λ4π₀. The first term is the usual Coulomb potential; the second term is a constant shift Δ U₀ = e² λ /(4π₀).
4. Compute Δ U₀: Δ U₀ = e² λ4π₀ = (e²4π₀ a₀) × (λ a₀) = (27.2 eV) × (1.37 × 10^-4) ≈ 3.7 × 10^-3 eV. Here e²/(4π₀ a₀) = 2 × 13.6 eV = 27.2 eV is twice the H ionisation energy.
5. Since the modification is a (nearly) constant positive shift in U over the region where the electron lives, the ground-state energy shifts up by roughly the same amount: Δ E₁ ≈ +3.7 × 10^-3 eV ≈ +3.7 meV. The atom is therefore less tightly bound, but by less than a ten-thousandth of the original binding energy.

Δ E₁ ≈ +3.7 meV; the modification raises the ground-state energy by about 4× 10^-3 eV.

Concept used. Inside the radius R₀ = 1 , Coulomb's law is replaced by a power-law F ∝ r^-ε with ε = 0.1. The Bohr ground-state radius for normal H is a₀ = 0.53 < R₀, so the electron lives entirely inside the modified-law region. We must redo the Bohr derivation with the new force law.

1. Write the modified force on the electron at distance r (r < R₀), with q₁ q₂ = e · e = e² and attraction directed inward: |F(r)| = e²4π₀ R₀²(R₀r)^ε = e² R₀^ε-24π₀· r^-ε. Let K = e² R₀^ε-2/(4π₀); then |F(r)| = K r^-ε.
2. Centripetal balance: m_e v²r = K r^-ε ⇒ m_e v² = K r^1-ε.
3. Bohr quantisation, n=1: m_e v r = ⇒ v = m_e r. Square: v² = ²/(m_e² r²), so m_e v² = ²/(m_e r²).
4. Equate the two expressions for m_e v²: ²m_e r² = K r^1-ε ⇒ r^3-ε = ²m_e K. Solve for r: r = (²m_e K)^1/(3-ε).
5. Plug in ε = 0.1, R₀ = 10^-10 m. Compute K: K = e² R₀^-1.94π₀. Use e²/(4π₀) = 2.307 × 10^-28 N m² and R₀^-1.9 = (10^-10)^-1.9 = 10¹⁹ m^-1.9: K = 2.307 × 10^-28 × 10¹⁹ = 2.307 × 10^-9 N m^0.1. Then ²m_e K = (1.055× 10^-34)² (9.11× 10^-31)(2.307× 10^-9) = 1.113 × 10^-682.102 × 10^-39 = 5.297 × 10^-30 m^2.9.
6. Take the 1/2.9 power: r = (5.297× 10^-30)^1/2.9 m. Compute the exponent in ₁₀ terms: ₁₀(5.297× 10^-30) = ₁₀ 5.297 - 30 = 0.724 - 30 = -29.276. Divide by 2.9: -29.276/2.9 = -10.095. So r = 10^-10.095 m = 0.804× 10^-10 m = 0.80 . (Consistent with r < R₀ = 1 : assumption holds.)
7. Velocity: v = m_e r = 1.055× 10^-34(9.11× 10^-31)(8.04× 10^-11) = 1.055× 10^-347.324× 10^-41 = 1.441 × 10⁶ m/s. Kinetic energy: aligned KE &= 12 m_e v² = 12 (9.11× 10^-31)(1.441× 10⁶)²
   &= 12 × 9.11 × 10^-31 × 2.077× 10¹²
   &= 9.461× 10^-19 J. aligned Convert: 9.461× 10^-19/1.6× 10^-19 = 5.91 eV.
8. Potential energy. Integrate F from r to R₀ to find U(r) - U(R₀) in the modified region; the value of U at R₀ is the usual Coulomb potential evaluated at R₀ (the two laws match continuously at r = R₀): U(R₀) = -e²4π₀ R₀ = -2.307× 10^-2810^-10 = -2.307× 10^-18 J = -14.4 eV. Inside: U(r) - U(R₀) = _r^R₀ K s^-ε(-1) ds·(-1); sign convention: F = -dU/dr, so U(r) = U(R₀) - _r^R₀ F(s) ds, with the attractive force having magnitude K s^-ε: U(r) = U(R₀) - _r^R₀ K s^-ε(-1) ds = U(R₀) + K _r^R₀ s^-ε ds. Actually, with attractive force (pointing toward origin), F_r = -K r^-ε (negative radial), and U(r) = -_∞^r F_r dr. For r0, (0.804)^{0.9}="10^{-9}\times" (1-\varepsilon)="K/0.9" (wait;="" +="" -="" 0.821="8.21\times" 1.79\times="" 10^{-10}="4.588" 10^{-10})^{0.9}="10^{-9}\times" 10^{-10}\).="" 10^{-19}\,\text{j}="2.87\,\text{eV}." 10^{-9}="" 10^{-9}\)="" 2.563\times="" 2.87="-11.53\,\)eV." 8.21\times="" 1-ε="0.9," ε="0.1:" k="" (r^0.9="(0.804" (r₀^0.9="" (u(r)="-14.4" ="" ="" k1-ε(r₀^1-ε="" _r^r₀="" ×="" and="" in="" k="" m(^0.9.="" makes="" modification="" potential="" r^{0.9}="10^{-9}" r^{1-\varepsilon}\right).="" s^{-\varepsilon}\,ds="U(R_0)" si.="" so="" the="" therefore="" u(r)="" u(r_0)="2.563\times" with="">less negative inside than the pure Coulomb, since the force grows more slowly than 1/r². So U(r) > U_Coulomb(r), as expected.)
9. Total energy: E = KE + U = 5.91 - 11.53 = -5.62 eV. So the binding energy in the modified theory is about 5.6 eV; substantially less than the standard 13.6 eV.

Ground-state radius r ≈ 0.80 ; ground-state energy E₁ ≈ -5.6 eV (compared with -13.6 eV in unmodified Coulomb law).