Atoms and Nuclei MCQs with Answers – Part 4 (NEET Physics)

Explanation: \(\textbf{Distinguish angular-momentum changes from energy changes:}\) The orbital angular momenta of adjacent Bohr states differ by: \[ \Delta L=(n+1)\hbar-n\hbar \] \[ \Delta L=\hbar \] However, the energies of successive Bohr states are not equally spaced. The absorbed photon energy must be obtained from the actual stationary-state energy difference. \(\textbf{Calculate the photon energy for P:}\) The transition is: \[ 2\to3 \] Therefore: \[ E_P=13.6\left(\frac{1}{4}-\frac{1}{9}\right) \] \[ E_P=13.6\left(\frac{5}{36}\right) \] \[ E_P=\frac{68}{36}\,\text{eV} \] \[ E_P\approx1.8889\,\text{eV} \] \(\textbf{Calculate the photon energy for Q:}\) The transition is: \[ 3\to4 \] Thus: \[ E_Q=13.6\left(\frac{1}{9}-\frac{1}{16}\right) \] \[ E_Q=13.6\left(\frac{7}{144}\right) \] \[ E_Q\approx0.6611\,\text{eV} \] \(\textbf{Compare the photon energies:}\) \[ \frac{E_P}{E_Q} =\frac{13.6(5/36)}{13.6(7/144)} \] Cancelling the common factor: \[ \frac{E_P}{E_Q}=\frac{5}{36}\times\frac{144}{7} \] \[ \frac{E_P}{E_Q}=\frac{20}{7} \] \(\textbf{Physical interpretation:}\) Both absorptions increase electron orbital angular momentum by the same quantum \(\hbar\). Nevertheless, the photon energies differ because the stationary-state energies depend on \(1/n^2\), rather than varying linearly with \(n\). Equal changes in angular momentum do not imply equal changes in total atomic energy. This is an important distinction between angular-momentum quantisation and the nonuniform spacing of hydrogen energy levels. \(\textbf{Final answer:}\) \[ \boxed{\frac{E_P}{E_Q}=\frac{20}{7}} \]

Explanation: \(\textbf{Identify the initial bound-state energy:}\) For hydrogen: \[ E_3=-\frac{13.6}{9} \] \[ E_3\approx-1.5111\,\text{eV} \] The initial orbital kinetic energy is: \[ K_3=-E_3 \] \[ K_3\approx1.5111\,\text{eV} \] \(\textbf{Use the Coulomb circular-orbit energy relation:}\) For the original circular orbit: \[ K_3=\frac{1}{2}m_ev_3^2 \] The potential energy satisfies: \[ U_3=-2K_3 \] Immediately after the interaction, the radius is unchanged. Therefore, the electrostatic potential energy remains \(U_3\) at that instant. \(\textbf{Determine the speed required for zero total energy:}\) For zero total mechanical energy: \[ K'+U_3=0 \] Therefore: \[ K'=-U_3 \] \[ K'=2K_3 \] Using kinetic energy: \[ \frac{1}{2}m_ev'^2=2\left(\frac{1}{2}m_ev_3^2\right) \] \[ v'^2=2v_3^2 \] \[ v'=\sqrt2\,v_3 \] \(\textbf{Calculate the energy supplied:}\) \[ W=K'-K_3 \] \[ W=2K_3-K_3 \] \[ W=K_3 \] \[ W\approx1.5111\,\text{eV} \] \(\textbf{Determine the new angular momentum:}\) The impulse is tangential and the radius is initially unchanged. Therefore: \[ L'=m_er_3v' \] \[ L'=\sqrt2\,m_er_3v_3 \] Since: \[ m_er_3v_3=3\hbar \] we obtain: \[ L'=3\sqrt2\,\hbar \] \(\textbf{Physical interpretation:}\) The supplied energy is precisely the ionisation energy from \(n=3\). However, the resulting motion is not another stationary circular Bohr orbit. The electron's total mechanical energy is zero, corresponding to the classical escape threshold. Its angular momentum need not equal an integer multiple of \(\hbar\) during this idealised nonstationary escape motion. Bohr's quantisation condition applies to the permitted bound circular stationary states, not to an electron escaping from the atom. \(\textbf{Final answer:}\) \[ \boxed{v'=\sqrt2\,v_3,\qquad W\approx1.511\,\text{eV},\qquad L'=3\sqrt2\,\hbar} \]

Explanation: \(\textbf{Determine the accessible excitation thresholds:}\) The ground-state energy is: \[ E_1=-13.6\,\text{eV} \] For excitation to \(n=3\): \[ \Delta E_{1\to3}=13.6\left(1-\frac{1}{9}\right) \] \[ \Delta E_{1\to3}=13.6\left(\frac{8}{9}\right) \] \[ \Delta E_{1\to3}\approx12.0889\,\text{eV} \] For excitation to \(n=4\): \[ \Delta E_{1\to4}=13.6\left(1-\frac{1}{16}\right) \] \[ \Delta E_{1\to4}=12.75\,\text{eV} \] \(\textbf{Test the incident electron's available energy:}\) The electron initially has: \[ K_i=12.40\,\text{eV} \] This exceeds the \(1\to3\) threshold but is below the \(1\to4\) threshold. Therefore, the highest accessible bound state by a single excitation collision is \(n=3\). \(\textbf{Calculate the outgoing electron energy:}\) If the atom undergoes \(1\to3\), the incident electron loses the excitation energy. Thus: \[ K_f=K_i-\Delta E_{1\to3} \] \[ K_f=12.40-12.0889 \] \[ K_f\approx0.3111\,\text{eV} \] \(\textbf{Assess the photon experiment:}\) A single photon is absorbed in a bound-bound transition only when its energy matches an allowed energy difference. The photon energy \(12.40\,\text{eV}\) does not equal either of the calculated excitation energies. It also remains below the ground-state ionisation threshold of \(13.6\,\text{eV}\). Therefore, ideal single-photon absorption cannot produce a transition from the ground state under the stated assumptions. \(\textbf{Physical interpretation:}\) An incident free electron can transfer only part of its kinetic energy during an inelastic collision. A photon, by contrast, is absorbed as a whole in the specified single-photon process. Consequently, equal incident energies do not guarantee equivalent excitation outcomes. The collision energy must be sufficient for the selected excitation, whereas photon absorption requires the appropriate resonance or ionisation threshold. \(\textbf{Final answer:}\) \[ \boxed{n_{\max}=3,\qquad K_{e,f}\approx0.311\,\text{eV}} \] The specified photon is not absorbed by ground-state hydrogen in the idealised model.

Explanation: \(\textbf{Determine the emitted photon's energy:}\) The first hydrogen atom undergoes: \[ 4\to3 \] The emitted energy is: \[ E_\gamma=13.6\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \] \[ E_\gamma=13.6\left(\frac{1}{9}-\frac{1}{16}\right) \] \[ E_\gamma=13.6\left(\frac{7}{144}\right) \] \[ E_\gamma\approx0.6611\,\text{eV} \] \(\textbf{Determine the ionisation condition:}\) For hydrogen initially in state \(n\): \[ I_n=\frac{13.6}{n^2}\,\text{eV} \] Ionisation is possible when: \[ E_\gamma\geq I_n \] Therefore: \[ 0.6611\geq\frac{13.6}{n^2} \] Rearranging: \[ n^2\geq\frac{13.6}{0.6611} \] \[ n^2\gtrsim20.57 \] \[ n\gtrsim4.54 \] \(\textbf{Apply the allowed-quantum-number restriction:}\) The principal quantum number must be a positive integer. Therefore, the smallest permitted value satisfying this condition is: \[ n_{\min}=5 \] \(\textbf{Check the neighbouring state:}\) For \(n=4\): \[ I_4=\frac{13.6}{16} \] \[ I_4=0.850\,\text{eV} \] The photon energy is insufficient to ionise an atom from that state. For \(n=5\): \[ I_5=\frac{13.6}{25} \] \[ I_5=0.544\,\text{eV} \] The threshold is now smaller than the incident photon energy. \(\textbf{Determine the liberated electron's kinetic energy:}\) \[ K_e=E_\gamma-I_5 \] \[ K_e=0.6611-0.544 \] \[ K_e\approx0.1171\,\text{eV} \] \(\textbf{Physical interpretation:}\) A photon emitted during a transition between two bound states can still ionise another hydrogen atom. The outcome depends on the initial binding energy of the absorbing atom. The photon is insufficient to ionise hydrogen in \(n=4\) but sufficient for hydrogen in \(n=5\). The smallest compatible integer quantum number follows from comparing the photon energy with the entire family of ionisation thresholds. \(\textbf{Final answer:}\) \[ \boxed{n_{\min}=5,\qquad K_e\approx0.117\,\text{eV}} \]

Explanation: The hydrogen-like energy formula describes one electron moving under the Coulomb attraction of a nucleus. Hydrogen itself is the standard example because its neutral atom contains one electron. Singly ionised helium also contains exactly one bound electron. Its nucleus has positive charge \(+2e\), so the elementary one-electron energy formula can be applied with \(Z=2\). Neutral helium, however, contains two bound electrons. Each electron experiences attraction from the nucleus as well as interaction with the other electron. The resulting electronic system cannot be described by simply treating both electrons as independent hydrogen-like particles in the unmodified nuclear Coulomb field. Therefore, using the one-electron formula with \(Z=2\) does not provide a valid complete prediction of neutral helium's energy levels and spectrum. The common nuclear charge does not eliminate the difference in electronic structure. The student has confused equality of nuclear charge with equality of the electronic Hamiltonian and allowed energy spectrum. The elementary Bohr model succeeds in explaining the main spectral features of hydrogen and can be extended to one-electron ions. Its limitations become important for systems containing more than one electron. The correct conclusion therefore distinguishes the range of applicability of the model rather than rejecting all ions or accepting every atom with a known atomic number.

Explanation: \(\textbf{Use the ionisation-energy ratio:}\) For hydrogen: \[ I_n=\frac{13.6}{n^2}\,\text{eV} \] Therefore: \[ \frac{I_P}{I_Q}=\frac{n_Q^2}{n_P^2} \] Substituting: \[ \frac{25}{9}=\frac{n_Q^2}{n_P^2} \] Taking positive square roots: \[ \frac{n_Q}{n_P}=\frac{5}{3} \] \(\textbf{Use the independent angular-momentum measurement:}\) Bohr's condition is: \[ L_n=n\hbar \] Thus: \[ L_Q-L_P=(n_Q-n_P)\hbar \] The measured difference is \(2\hbar\). Therefore: \[ n_Q-n_P=2 \] \(\textbf{Solve both conditions simultaneously:}\) From the ratio: \[ n_Q=\frac{5}{3}n_P \] Substituting into the difference: \[ \frac{5}{3}n_P-n_P=2 \] \[ \frac{2}{3}n_P=2 \] \[ n_P=3 \] Therefore: \[ n_Q=5 \] \(\textbf{Calculate the required absorption energy:}\) The upward transition is: \[ 3\to5 \] Its energy requirement is: \[ E_{\text{abs}}=13.6\left(\frac{1}{3^2}-\frac{1}{5^2}\right) \] \[ E_{\text{abs}}=13.6\left(\frac{1}{9}-\frac{1}{25}\right) \] \[ E_{\text{abs}}=13.6\left(\frac{16}{225}\right) \] \[ E_{\text{abs}}\approx0.9671\,\text{eV} \] \(\textbf{Physical interpretation:}\) The ratio of ionisation energies determines the ratio of the principal quantum numbers but not their absolute values. The angular-momentum difference provides an independent integer constraint. Together, the measurements uniquely identify the two permitted states. The absorbed photon energy is then obtained from their signed stationary-state energies. This demonstrates why a ratio measurement and a difference measurement can contain more information together than either measurement alone. \(\textbf{Final answer:}\) \[ \boxed{n_P=3,\qquad n_Q=5,\qquad E_{\text{abs}}\approx0.967\,\text{eV}} \]

Explanation: For a permitted circular Bohr orbit, the electron revolution frequency is related to its kinetic energy by \(hf_n=2K_n/n\). For hydrogen, the orbital kinetic energy \(K_n\) equals the ionisation energy \(I_n\) from that state. Therefore, a photon having frequency \(f_n\) possesses energy \(2I_n/n\). For the third orbit, this is only two-thirds of the ionisation energy. Statement I is therefore correct. Statement II is incorrect because the mechanical revolution frequency does not automatically equal an allowed atomic transition frequency. The latter is determined by the energy difference between two stationary states divided by Planck's constant. A photon with a particular revolution-frequency value must still satisfy the appropriate transition-energy condition before resonant bound-bound absorption is possible. Statement III correctly distinguishes these two different physical relations. Statement IV is also correct because the photon carries less energy than the ionisation threshold from \(n=3\). It therefore cannot remove the electron through the specified single-photon process. The absence of sufficient photon energy is decisive even before considering possible bound-bound resonances. The analysis also reinforces that an electron does not continuously radiate at its revolution frequency while remaining in a stationary Bohr state. Thus, Statements I, III and IV are correct.

Explanation: \(\textbf{Identify the initial and final energies:}\) The atom begins in P: \[ E_P=-0.85\,\text{eV} \] It eventually reaches R: \[ E_R=-13.60\,\text{eV} \] The total energy released is determined by these endpoints. \(\textbf{Calculate the first photon energy:}\) For P \(\to\) Q: \[ E_{\gamma1}=E_P-E_Q \] \[ E_{\gamma1}=-0.85-(-3.40) \] \[ E_{\gamma1}=2.55\,\text{eV} \] \(\textbf{Calculate the second photon energy:}\) For Q \(\to\) R: \[ E_{\gamma2}=E_Q-E_R \] \[ E_{\gamma2}=-3.40-(-13.60) \] \[ E_{\gamma2}=10.20\,\text{eV} \] \(\textbf{Calculate total emitted energy:}\) \[ E_{\text{total}}=E_{\gamma1}+E_{\gamma2} \] \[ E_{\text{total}}=2.55+10.20 \] \[ E_{\text{total}}=12.75\,\text{eV} \] \(\textbf{Apply the detector's selection condition:}\) The detector counts only photons having: \[ E_\gamma\gt 5.0\,\text{eV} \] The first photon carries \(2.55\,\text{eV}\), so it is not counted. The second photon carries \(10.20\,\text{eV}\), so it is counted. Therefore: \[ N_{\text{detected}}=1 \] The energy recorded in this ideal detector is: \[ E_{\text{recorded}}=10.20\,\text{eV} \] \(\textbf{Endpoint-energy check:}\) The total emitted energy must also satisfy: \[ E_{\text{total}}=E_P-E_R \] \[ E_{\text{total}}=-0.85-(-13.60) \] \[ E_{\text{total}}=12.75\,\text{eV} \] \(\textbf{Physical interpretation:}\) The atom emits two photons, but the detector records only one because its response is restricted by photon energy. The rejected photon still carries away energy and contributes to the atom's complete energy balance. Thus, detected photon energy must not be confused with total emitted radiation energy. The endpoint-energy calculation provides an independent check on the cascade. \(\textbf{Final answer:}\) \[ \boxed{(E_{\text{total}},N_{\text{detected}},E_{\text{recorded}})=(12.75\,\text{eV},1,10.20\,\text{eV})} \]

Explanation: For hydrogen, the electron's orbital kinetic energy decreases as the principal quantum number increases. This follows from the relation \(K_n=13.6/n^2\,\text{eV}\). Therefore, Statement I is correct. The de Broglie wavelength of a moving non-relativistic electron is \(h/p\). Its kinetic energy is \(p^2/(2m_e)\). Multiplying the kinetic energy by the square of the wavelength eliminates the momentum dependence. The resulting product is \(h^2/(2m_e)\), which is independent of the electron's speed and the particular permitted orbit. Statement II is consequently correct. However, the relationship between kinetic energy and wavelength squared is inverse, not directly proportional. As wavelength increases, kinetic energy decreases. Thus, Graph Q is not a positive-slope straight line through the origin, making Statement III incorrect. Statement IV is correct because the constant product follows from general non-relativistic matter-wave kinematics. Its validity is not restricted to bound atomic stationary states. Bohr quantisation additionally requires the integral standing-wave condition and Coulomb circular-force balance. Therefore, observing the constant product alone would not establish that an electron occupies a permitted Bohr orbit. The correct combination is I, II and IV.

Explanation: \(\textbf{Identify the original Coulomb energy relation:}\) For a circular Coulomb orbit: \[ U=-2K \] Therefore, the original total mechanical energy is: \[ E=K+U \] \[ E=-K \] \(\textbf{Apply the instantaneous change:}\) The electron's new kinetic energy is: \[ K'=\frac{3K}{2} \] The interaction is specified to leave its radial position unchanged. Since Coulomb potential energy depends only on radial separation: \[ U'=U \] \[ U'=-2K \] \(\textbf{Calculate the new total mechanical energy:}\) \[ E'=K'+U' \] \[ E'=\frac{3K}{2}-2K \] \[ E'=-\frac{K}{2} \] The total energy remains negative. Thus, the instantaneous state has not reached the zero-energy escape threshold. \(\textbf{Test the circular-force condition:}\) The new kinetic energy is greater than the original kinetic energy. Because: \[ K=\frac{1}{2}m_ev^2 \] the new speed is: \[ v'=\sqrt{\frac{3}{2}}\,v \] At the unchanged radius, the centripetal force required for circular motion would be: \[ F'_{\text{required}}=\frac{m_ev'^2}{r} \] Therefore: \[ F'_{\text{required}}=\frac{3}{2}\frac{m_ev^2}{r} \] The actual Coulomb force has not changed because the radius and nuclear charge remain unchanged. It therefore cannot supply the new required centripetal force. \(\textbf{Physical interpretation:}\) Negative total mechanical energy does not automatically mean that an electron occupies a stationary circular Bohr orbit. The disturbed electron still has negative total energy, but its instantaneous speed is incompatible with circular force balance at the unchanged radius. This separates the classical criterion for a bound Coulomb trajectory from the stricter stationary-orbit conditions of the Bohr model. \(\textbf{Final answer:}\) \[ \boxed{E'=-\frac{K}{2},\qquad v'=\sqrt{\frac{3}{2}}\,v} \] The disturbed motion is not the original stationary circular Bohr orbit.

Explanation: \(\textbf{Use the Coulomb circular-orbit energy relation:}\) For a one-electron ion having nuclear charge \(+Ze\), the electron's orbital kinetic energy is: \[ K=\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{2r} \] Writing \(k=1/(4\pi\varepsilon_0)\): \[ K=\frac{kZe^2}{2r} \] \(\textbf{Compare the two measured states:}\) The measured radii satisfy: \[ r_P=r_Q \] The measured kinetic energies satisfy: \[ K_P=K_Q \] Therefore: \[ \frac{kZ_Pe^2}{2r_P}=\frac{kZ_Qe^2}{2r_Q} \] Since the radii are equal: \[ Z_P=Z_Q \] \(\textbf{Apply the Bohr radius relation:}\) For a one-electron ion: \[ r_n=\frac{n^2a_0}{Z} \] The nuclear charge numbers are now known to be equal. The orbital radii are also equal. Therefore: \[ \frac{n_P^2a_0}{Z_P}=\frac{n_Q^2a_0}{Z_Q} \] Thus: \[ n_P^2=n_Q^2 \] Because principal quantum numbers are positive integers: \[ n_P=n_Q \] \(\textbf{Physical interpretation:}\) Equality of orbital radius alone would not establish equality of nuclear charge or orbit number. Different combinations of \(n\) and \(Z\) can produce the same radius. However, the additional kinetic-energy measurement removes that freedom because kinetic energy at a fixed Coulomb radius is directly proportional to \(Z\). Once nuclear charge is fixed, equal permitted radii imply equal principal quantum numbers. The two measurements together therefore rule out the student's proposal. \(\textbf{Final answer:}\) \[ Z_P=Z_Q,\qquad n_P=n_Q \]

Explanation: \(\textbf{Identify the initial energy reference:}\) The electron is initially very far from the proton. The electrostatic potential energy is therefore taken as zero. Its initial kinetic energy is: \[ K_i=1.50\,\text{eV} \] Thus, the initial energy of the separated electron–proton system, relative to the stated reference, is: \[ E_i=+1.50\,\text{eV} \] \(\textbf{Determine the final bound-state energy:}\) The electron is captured into: \[ n=2 \] The corresponding total energy is: \[ E_f=-\frac{13.6}{4} \] \[ E_f=-3.40\,\text{eV} \] \(\textbf{Apply energy conservation:}\) The final system contains a bound hydrogen atom and the emitted photon. Neglecting recoil: \[ E_i=E_f+E_\gamma \] Therefore: \[ E_\gamma=E_i-E_f \] Substituting: \[ E_\gamma=1.50-(-3.40) \] \[ E_\gamma=4.90\,\text{eV} \] \(\textbf{Determine the photon wavelength:}\) \[ E_\gamma=\frac{hc}{\lambda} \] Therefore: \[ \lambda=\frac{hc}{E_\gamma} \] \[ \lambda=\frac{1240\,\text{eV\,nm}}{4.90\,\text{eV}} \] \[ \lambda\approx253.1\,\text{nm} \] \(\textbf{Compare with the ionisation threshold:}\) The ionisation energy from \(n=2\) is: \[ I_2=3.40\,\text{eV} \] Therefore: \[ E_\gamma-I_2=4.90-3.40 \] \[ E_\gamma-I_2=1.50\,\text{eV} \] \(\textbf{Physical interpretation:}\) The emitted photon carries both the energy released by binding the electron in the second state and the kinetic energy initially possessed by the free electron. It therefore has greater energy than the threshold photon needed to ionise hydrogen from \(n=2\). This is the energy-conservation reverse of photoionisation with an outgoing electron possessing \(1.50\,\text{eV}\) of kinetic energy. The photon is associated with capture from the continuum rather than a transition between two bound Bohr levels. Consequently, it should not be classified as an ordinary bound-bound Balmer spectral line merely because the final electron state is \(n=2\). \(\textbf{Final answer:}\) \[ \boxed{E_\gamma=4.90\,\text{eV},\qquad\lambda\approx253\,\text{nm}} \]

Explanation: \(\textbf{Distinguish adjacent excitation from ionisation:}\) The ionisation energy from state \(n\) is: \[ I_n=\frac{13.6}{n^2}\,\text{eV} \] The energy required for the adjacent upward transition is: \[ \Delta E_n=13.6\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\,\text{eV} \] \(\textbf{Form the relevant energy ratio:}\) Dividing the excitation energy by the ionisation energy: \[ \frac{\Delta E_n}{I_n}=1-\frac{n^2}{(n+1)^2} \] Simplifying: \[ \frac{\Delta E_n}{I_n}=\frac{2n+1}{(n+1)^2} \] The required condition becomes: \[ \frac{2n+1}{(n+1)^2}\lt \frac{1}{2} \] Cross-multiplying the positive denominator: \[ 2(2n+1)\lt (n+1)^2 \] \[ 4n+2\lt n^2+2n+1 \] \[ n^2-2n-1\gt 0 \] \(\textbf{Identify the smallest permitted state:}\) For \(n=2\): \[ n^2-2n-1=4-4-1=-1 \] The condition is not satisfied. For \(n=3\): \[ n^2-2n-1=9-6-1=2 \] The condition is satisfied. Therefore: \[ n_{\min}=3 \] \(\textbf{Calculate the corresponding excitation energy:}\) The relevant transition is: \[ 3\to4 \] \[ \Delta E_3=13.6\left(\frac{1}{9}-\frac{1}{16}\right) \] \[ \Delta E_3=13.6\left(\frac{7}{144}\right) \] \[ \Delta E_3\approx0.6611\,\text{eV} \] \(\textbf{Determine the photon wavelength:}\) \[ \lambda=\frac{hc}{\Delta E_3} \] \[ \lambda=\frac{1240}{0.6611}\,\text{nm} \] \[ \lambda\approx1876\,\text{nm} \] \(\textbf{Physical interpretation:}\) The energy required for the next adjacent excitation decreases more rapidly than the remaining ionisation energy as the principal quantum number increases. Beginning at \(n=3\), the next excitation requires less than half the energy needed for complete ionisation. This comparison requires two different energy differences: one between neighbouring bound states and another between a bound state and the continuum. The photon wavelength is determined by the adjacent excitation energy, not by the ionisation threshold. \(\textbf{Final answer:}\) \[ \boxed{n_{\min}=3,\qquad\lambda_{3\to4}\approx1876\,\text{nm}} \]

Explanation: \(\textbf{Separate photon count from energy release:}\) Every atom begins in \(n=4\) and eventually reaches \(n=1\). The number of emitted photons depends on the pathway. However, the total energy released by each atom depends only on its initial and final energy levels. \(\textbf{Calculate the pathway populations:}\) The preparation rate is: \[ R=1.0\times10^{12}\,\text{s}^{-1} \] For pathway P: \[ R_P=0.40R \] \[ R_P=4.0\times10^{11}\,\text{s}^{-1} \] For pathway Q: \[ R_Q=0.60R \] \[ R_Q=6.0\times10^{11}\,\text{s}^{-1} \] \(\textbf{Determine the total photon rate:}\) Each atom following P emits two photons. Each atom following Q emits three photons. Therefore: \[ R_\gamma=2R_P+3R_Q \] \[ R_\gamma=2(4.0\times10^{11})+3(6.0\times10^{11}) \] \[ R_\gamma=8.0\times10^{11}+1.8\times10^{12} \] \[ R_\gamma=2.6\times10^{12}\,\text{s}^{-1} \] \(\textbf{Identify the selected spectral line:}\) Only pathway Q contains the transition \(4\to3\). Each Q atom produces one such photon. Thus: \[ R_{4\to3}=R_Q \] \[ R_{4\to3}=6.0\times10^{11}\,\text{s}^{-1} \] \(\textbf{Calculate the energy released by each atom:}\) For hydrogen: \[ E_4=-\frac{13.6}{16} \] \[ E_4=-0.85\,\text{eV} \] The ground-state energy is: \[ E_1=-13.6\,\text{eV} \] Therefore: \[ E_{\text{released}}=E_4-E_1 \] \[ E_{\text{released}}=-0.85-(-13.6) \] \[ E_{\text{released}}=12.75\,\text{eV} \] \(\textbf{Calculate the total emitted power:}\) \[ P=R E_{\text{released}} \] Converting the energy per atom into joules: \[ P=(1.0\times10^{12})(12.75)(1.60\times10^{-19}) \] \[ P=2.04\times10^{-6}\,\text{W} \] \[ P=2.04\,\mu\text{W} \] \(\textbf{Physical interpretation:}\) The branching fractions control how many photons are emitted and which spectral lines appear. They do not change the total energy released by an atom returning between the same initial and final states. Consequently, the overall photon rate depends on the mixture of pathways, while the total emitted power depends on the common endpoint-energy difference. The rate for one particular spectral line must be calculated using only the pathways containing that transition. \(\textbf{Final answer:}\) \[ \boxed{(R_\gamma,R_{4\to3},P)=(2.6\times10^{12}\,\text{s}^{-1},6.0\times10^{11}\,\text{s}^{-1},2.04\,\mu\text{W})} \]

Explanation: Observation P concerns the distribution of alpha-particle scattering angles in Rutherford's experiment. The passage of most alpha particles through a thin foil shows that the atom contains a large region through which the projectiles can pass without strong deflection. The rare large-angle deflections require strong electrostatic repulsion concentrated within a very small region. This supports the nuclear model of the atom, giving P–2. Observation Q concerns the discrete frequencies of photons emitted by hydrogen. Bohr's model assigns definite energies to permitted stationary states. A transition between two such states produces a photon whose energy equals their energy difference. Therefore, Q matches statement 4. Observation R concerns the spacing of bound-state energies. For hydrogen, the energy magnitude decreases as the inverse square of principal quantum number. At large \(n\), the negative energies approach zero, which represents the ionisation limit under the usual reference convention. Thus, R matches statement 3. Observation S concerns the relative intensities of different spectral lines. Knowing the energy differences between states determines the corresponding photon frequencies but does not, by itself, determine how frequently different transitions occur. Therefore, S matches statement 1. The observations test different explanatory roles: evidence for the nuclear structure, the quantisation of atomic energy and the limitations of the elementary atomic model. The correct matching is P–2, Q–4, R–3 and S–1.

Explanation: \(\textbf{Determine the receiving atoms' ionisation threshold:}\) Every receiving hydrogen atom initially occupies \(n=2\). Therefore: \[ I_2=\frac{13.6}{2^2} \] \[ I_2=3.40\,\text{eV} \] A photon can ionise such an atom only if its energy is at least \(3.40\,\text{eV}\). \(\textbf{Analyse direct emission in pathway P:}\) The donor undergoes: \[ 4\to1 \] The emitted photon energy is: \[ E_{\gamma,P}=13.6\left(1-\frac{1}{16}\right) \] \[ E_{\gamma,P}=12.75\,\text{eV} \] The receiving atom can be ionised. Its photoelectron kinetic energy is: \[ K_P=E_{\gamma,P}-I_2 \] \[ K_P=12.75-3.40 \] \[ K_P=9.35\,\text{eV} \] \(\textbf{Analyse the first photon from pathway Q:}\) The first transition is: \[ 4\to3 \] Its photon energy is: \[ E_{\gamma,Q1}=13.6\left(\frac{1}{9}-\frac{1}{16}\right) \] \[ E_{\gamma,Q1}\approx0.6611\,\text{eV} \] This is less than the receiving atom's ionisation threshold. Therefore, the first photon cannot ionise its receiver. \(\textbf{Analyse the second photon from pathway Q:}\) The second transition is: \[ 3\to1 \] Its photon energy is: \[ E_{\gamma,Q2}=13.6\left(1-\frac{1}{9}\right) \] \[ E_{\gamma,Q2}\approx12.0889\,\text{eV} \] This exceeds the ionisation threshold. The corresponding photoelectron kinetic energy is: \[ K_Q=E_{\gamma,Q2}-I_2 \] \[ K_Q=12.0889-3.40 \] \[ K_Q\approx8.6889\,\text{eV} \] \(\textbf{Compare the pathways:}\) Both pathways release the same total energy from the donor atom: \[ 12.75\,\text{eV} \] However, Q distributes that energy between two unequal photons. Only its higher-energy photon can ionise a receiving atom prepared in \(n=2\). Consequently, equal total emitted radiation energy does not guarantee equal photoelectron kinetic energies or equal ionisation outcomes. \(\textbf{Final answer:}\) \[ \boxed{K_P=9.35\,\text{eV},\qquad K_Q\approx8.69\,\text{eV}} \]

Explanation: \(\textbf{Determine the energy emitted by the donor:}\) The transition is: \[ n\to3 \] Its photon energy is: \[ E_\gamma=13.6\left(\frac{1}{9}-\frac{1}{n^2}\right)\,\text{eV} \] \(\textbf{Determine the receiver's binding energy:}\) The second hydrogen atom initially occupies the same state \(n\). Its ionisation energy is: \[ I_n=\frac{13.6}{n^2}\,\text{eV} \] \(\textbf{Form the ionisation condition:}\) Single-photon ionisation requires: \[ E_\gamma\geq I_n \] Therefore: \[ 13.6\left(\frac{1}{9}-\frac{1}{n^2}\right)\geq\frac{13.6}{n^2} \] Cancelling \(13.6\): \[ \frac{1}{9}-\frac{1}{n^2}\geq\frac{1}{n^2} \] \[ \frac{1}{9}\geq\frac{2}{n^2} \] \[ n^2\geq18 \] \(\textbf{Apply the principal-quantum-number restriction:}\) The donor must have \(n\gt 3\). The smallest integer satisfying: \[ n^2\geq18 \] is: \[ n_{\min}=5 \] \(\textbf{Calculate the emitted photon energy for this case:}\) For the transition \(5\to3\): \[ E_\gamma=13.6\left(\frac{1}{9}-\frac{1}{25}\right) \] \[ E_\gamma=13.6\left(\frac{16}{225}\right) \] \[ E_\gamma\approx0.9671\,\text{eV} \] \(\textbf{Determine the receiving atom's ionisation threshold:}\) \[ I_5=\frac{13.6}{25} \] \[ I_5=0.544\,\text{eV} \] \(\textbf{Calculate the photoelectron kinetic energy:}\) \[ K_e=E_\gamma-I_5 \] \[ K_e=0.9671-0.544 \] \[ K_e\approx0.4231\,\text{eV} \] \(\textbf{Physical interpretation:}\) A photon emitted through a bound-bound transition can possess enough energy to ionise another atom initially in a sufficiently excited state. For \(n=4\), the emitted \(4\to3\) photon has energy smaller than the receiving atom's ionisation threshold. At \(n=5\), the inequality reverses. The required minimum quantum number is established by combining the emission energy with the receiver's independent ionisation requirement. \(\textbf{Final answer:}\) \[ \boxed{n_{\min}=5,\qquad K_e\approx0.423\,\text{eV}} \]

Explanation: \(\textbf{Recover the initial state from its binding energy:}\) For hydrogen: \[ I_n=\frac{13.6}{n^2}\,\text{eV} \] The initial ionisation energy is: \[ I_i=0.850\,\text{eV} \] Therefore: \[ 0.850=\frac{13.6}{n_i^2} \] \[ n_i^2=16 \] \[ n_i=4 \] \(\textbf{Recover the final state:}\) The final ionisation energy is: \[ I_f=3.40\,\text{eV} \] Thus: \[ 3.40=\frac{13.6}{n_f^2} \] \[ n_f^2=4 \] \[ n_f=2 \] The atom undergoes the downward transition: \[ 4\to2 \] \(\textbf{Determine the emitted photon energy:}\) The total atomic energy is the negative of the ionisation energy. Therefore: \[ E_i=-0.850\,\text{eV} \] \[ E_f=-3.40\,\text{eV} \] For photon emission: \[ E_\gamma=E_i-E_f \] \[ E_\gamma=-0.850-(-3.40) \] \[ E_\gamma=2.55\,\text{eV} \] \(\textbf{Determine the photon wavelength:}\) \[ \lambda=\frac{hc}{E_\gamma} \] \[ \lambda=\frac{1240}{2.55}\,\text{nm} \] \[ \lambda\approx486.3\,\text{nm} \] \(\textbf{Compare the angular momenta:}\) Bohr's angular-momentum condition gives: \[ L_n=n\hbar \] Therefore: \[ \frac{L_f}{L_i}=\frac{n_f}{n_i} \] \[ \frac{L_f}{L_i}=\frac{2}{4} \] \[ \frac{L_f}{L_i}=\frac{1}{2} \] \(\textbf{Physical interpretation:}\) The ionisation energy increases because the atom ends in a more tightly bound state. Its signed total energy becomes more negative. The released photon carries the difference between the two binding-energy magnitudes. The orbital angular momentum decreases according to the change in principal quantum number. The emitted photon is the \(4\to2\) Balmer-series line. \(\textbf{Final answer:}\) \[ \boxed{4\to2,\qquad\lambda\approx486\,\text{nm},\qquad\frac{L_f}{L_i}=\frac{1}{2}} \]

Explanation: Hydrogen's bound-state energies are quantised according to the principal quantum number. For an atom initially in \(n=3\), the energy required for excitation to \(n=4\) is approximately \(0.6611\,\text{eV}\). Statement I is correct within the numerical precision specified. Statement II is incorrect because bound-bound photon absorption requires an allowed energy difference, not merely any energy greater than the first excitation threshold. Photons whose energies lie between two permitted transition energies are not resonantly absorbed under the idealised conditions. Statement III is correct because transitions from \(n=3\) to successively higher finite states have discrete, increasing excitation energies. These energies approach the ionisation threshold as the final principal quantum number increases without bound. The limiting photon energy is the energy required to move the electron to the continuum threshold. Statement IV is also correct because a photon carrying sufficient energy can ionise the atom. When recoil is neglected, energy exceeding the ionisation requirement becomes kinetic energy of the escaping electron. The ionisation continuum differs from the discrete collection of bound-bound transitions. The important distinction is between photons capable of producing a specific bound-state excitation and photons energetic enough to remove the electron altogether. Thus, Statements I, III and IV are correct.

Explanation: \(\textbf{Identify the collision's energy constraint:}\) The target atom initially occupies: \[ n_i=3 \] Its total energy is: \[ E_3=-\frac{13.6}{9} \] \[ E_3\approx-1.5111\,\text{eV} \] The incident electron has: \[ K_i=2.00\,\text{eV} \] The energy needed to ionise the target from \(n=3\) is: \[ I_3=1.5111\,\text{eV} \] \(\textbf{Compare with the bound-state excitation limit:}\) Since the incident electron has more than the ionisation threshold, it has enough energy in principle to reach every finite higher bound level, subject to the stated collision mechanism. The permitted bound-state energies approach zero from below as \(n_f\) increases. Therefore, there is no highest finite bound principal quantum number fixed by the stated \(2.00\,\text{eV}\) incident energy. \(\textbf{Check the proposed finite upper states:}\) The excitation energy for \(3\to7\) is: \[ \Delta E_{3\to7}=13.6\left(\frac{1}{9}-\frac{1}{49}\right) \] \[ \Delta E_{3\to7}=13.6\left(\frac{40}{441}\right) \] \[ \Delta E_{3\to7}\approx1.2336\,\text{eV} \] The outgoing electron energy would be: \[ K_f=2.00-1.2336 \] \[ K_f\approx0.7664\,\text{eV} \] However, excitation to \(n=8\), \(n=9\) and every other higher finite bound state would also require less than the ionisation energy \(1.5111\,\text{eV}\). Thus, none of the listed finite states is an upper bound on accessible bound states. \(\textbf{Physical interpretation:}\) Individual finite excitation states are accessible, but none is the highest accessible bound state. Once the collision energy exceeds the initial state's ionisation energy, the ideal Bohr-level model places no finite upper bound on the final bound-state quantum number. The incident electron could also possess enough energy for ionisation, although the stem restricts consideration to bound-bound excitation. The limiting outgoing electron kinetic energy is the initial kinetic energy minus the ionisation threshold: \[ K_{f,\mathrm{limit}}=2.00-1.5111 \] \[ K_{f,\mathrm{limit}}\approx0.4889\,\text{eV} \] For a bound level with finite \(n_f\), the required excitation energy remains strictly below the ionisation threshold, so the outgoing electron retains slightly more than this limiting kinetic energy. \(\textbf{Final answer:}\) \[ n_f\ \text{has no finite maximum},\qquad K_f\to0.489\,\text{eV}\ \text{from above} \]