Atomic Structure MCQs with Answers – Part 5 (NEET Chemistry)

Explanation: \(\textbf{Compare the Bohr orbital radii:}\) For hydrogen, the orbital radius is proportional to the square of the principal quantum number: \[ r_n=a_0n^2 \] Therefore, \[ \frac{r_4}{r_2}=\frac{4^2}{2^2} \] \[ \frac{r_4}{r_2}=4 \] The fourth orbit has four times the radius and circumference of the second orbit. \(\textbf{Apply the standing-wave condition:}\) The permitted circumference contains an integral number of electron wavelengths: \[ 2\pi r_n=n\lambda_n \] Rearranging, \[ \lambda_n=\frac{2\pi r_n}{n} \] Hence, \[ \frac{\lambda_4}{\lambda_2} =\frac{r_4}{r_2}\frac{2}{4} \] Substituting the radius ratio, \[ \frac{\lambda_4}{\lambda_2}=4\left(\frac12\right) \] \[ \frac{\lambda_4}{\lambda_2}=2 \] \(\textbf{Determine the momentum ratio:}\) The de Broglie relation is \[ \lambda=\frac hp \] Therefore, \[ \frac{p_4}{p_2}=\frac{\lambda_2}{\lambda_4} \] \[ \frac{p_4}{p_2}=\frac12 \] \(\textbf{Interpret the linked changes:}\) The larger orbit has a greater circumference, but it must accommodate more complete electron wavelengths. Consequently, the wavelength increases only twofold while the circumference increases fourfold. The electron momentum decreases because momentum is inversely proportional to de Broglie wavelength. The result also agrees with the Bohr speed relation \(v_n\propto1/n\) for fixed nuclear charge. The standing-wave condition provides a wave-based interpretation of angular-momentum quantisation rather than evidence for an exactly known classical electron trajectory. \(\textbf{Final answer:}\) \[ \boxed{\frac{r_4}{r_2}=4,\qquad\frac{\lambda_4}{\lambda_2}=2,\qquad\frac{p_4}{p_2}=\frac12} \]

Explanation: \(\textbf{Compare the momenta using wavelength:}\) For a photon, \[ p_\gamma=\frac h\lambda \] For a material particle, \[ p_e=\frac h\lambda \] Because their wavelengths are identical, \[ p_\gamma=p_e \] This equality concerns momentum magnitude, not particle speed or energy. \(\textbf{Calculate the photon energy:}\) The photon energy is \[ E_\gamma=\frac{hc}{\lambda} \] Substituting, \[ E_\gamma=\frac{1240\,\mathrm{eV\,nm}}{0.100\,\mathrm{nm}} \] \[ E_\gamma=12400\,\mathrm{eV} \] \[ E_\gamma=12.4\,\mathrm{keV} \] \(\textbf{Calculate the electron kinetic energy independently:}\) The supplied electron relation is \[ \lambda_e=\frac{1.227}{\sqrt{K_e}}\,\mathrm{nm} \] where \(K_e\) is expressed in electronvolts. Rearranging, \[ K_e=\left(\frac{1.227}{\lambda_e(\mathrm{nm})}\right)^2\mathrm{eV} \] Substituting, \[ K_e=\left(\frac{1.227}{0.100}\right)^2\mathrm{eV} \] \[ K_e=(12.27)^2\,\mathrm{eV} \] \[ K_e\approx150.6\,\mathrm{eV} \] \(\textbf{Interpret the energy difference:}\) The photon obeys the energy-momentum relation \(E_\gamma=pc\). The non-relativistic electron obeys \(K_e=p^2/(2m_e)\). These different relations produce different energies even when the momentum magnitudes are equal. A photon does not acquire an electron's mass merely because their wavelengths coincide. Likewise, a material particle does not move at the speed of light simply because it exhibits wave behaviour. The calculation demonstrates why wavelength equality must not be confused with energy equality. \(\textbf{Final answer:}\) \[ \boxed{p_\gamma=p_e,\qquad E_\gamma=12.4\,\mathrm{keV},\qquad K_e\approx151\,\mathrm{eV}} \]

Explanation: \(\textbf{Analyse Experiment P at equal kinetic energy:}\) For a non-relativistic particle, \[ K=\frac{p^2}{2m} \] Using the de Broglie relation, \[ \lambda=\frac h{\sqrt{2mK}} \] At equal kinetic energies, \[ \lambda\propto\frac1{\sqrt m} \] Therefore, \[ \frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_p}{m_e}} \] Using the given mass ratio, \[ \frac{\lambda_e}{\lambda_p}=\sqrt{1836} \] \[ \frac{\lambda_e}{\lambda_p}\approx42.8 \] The lighter electron has the greater wavelength at equal kinetic energy. \(\textbf{Analyse Experiment Q at equal wavelength:}\) The de Broglie equation gives \[ p=\frac h\lambda \] Equal wavelengths therefore imply equal momentum magnitudes. For fixed momentum, \[ K=\frac{p^2}{2m} \] Thus, \[ K\propto\frac1m \] Consequently, \[ \frac{K_e}{K_p}=\frac{m_p}{m_e} \] \[ \frac{K_e}{K_p}=1836 \] \(\textbf{Explain the reversal of the comparison:}\) In Experiment P, kinetic energy is held constant and momentum adjusts according to particle mass. In Experiment Q, momentum is held constant and kinetic energy adjusts according to particle mass. The proportionality governing wavelength in the first experiment consequently differs from the proportionality governing kinetic energy in the second. The electron's greater wavelength at equal kinetic energy does not imply that it must also have greater wavelength at equal momentum. Equal momentum requires equal de Broglie wavelength, irrespective of the particle masses. The two experimental conditions must not be interchanged while making a comparison. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{\lambda_e}{\lambda_p}\right)_{K\ \mathrm{equal}}=\sqrt{1836},\qquad\left(\frac{K_e}{K_p}\right)_{\lambda\ \mathrm{equal}}=1836} \]

Explanation: \(\textbf{Apply the uncertainty inequality:}\) For corresponding position and momentum components, \[ \Delta x\,\Delta p_x\geq\frac{\hbar}{2} \] The product must be at least \(\hbar/2\). It need not equal the lower bound in every physically permitted state. \(\textbf{Test measurement P:}\) \[ (\Delta x\,\Delta p_x)_P=a\left(\frac{\hbar}{3a}\right) \] \[ (\Delta x\,\Delta p_x)_P=\frac{\hbar}{3} \] Since \[ \frac{\hbar}{3}\lt \frac{\hbar}{2} \] P is incompatible with the uncertainty inequality. \(\textbf{Test measurement Q:}\) \[ (\Delta x\,\Delta p_x)_Q=2a\left(\frac{\hbar}{4a}\right) \] \[ (\Delta x\,\Delta p_x)_Q=\frac{\hbar}{2} \] Q satisfies the limiting equality and is compatible with the inequality. \(\textbf{Test measurement R:}\) \[ (\Delta x\,\Delta p_x)_R=\frac a4\left(\frac{2\hbar}{a}\right) \] \[ (\Delta x\,\Delta p_x)_R=\frac{\hbar}{2} \] R also satisfies the limiting equality. \(\textbf{Test measurement S:}\) \[ (\Delta x\,\Delta p_x)_S=\frac a2\left(\frac{\hbar}{2a}\right) \] \[ (\Delta x\,\Delta p_x)_S=\frac{\hbar}{4} \] Because \[ \frac{\hbar}{4}\lt \frac{\hbar}{2} \] S is incompatible with the uncertainty inequality. \(\textbf{Interpret the measurement proposals:}\) The inequality sets a fundamental lower limit on the product of the uncertainties. Reducing position uncertainty requires sufficiently large uncertainty in the corresponding momentum component. A measurement proposal with a smaller product cannot become valid merely through improved instrumental precision. The inequality does not require the position and momentum uncertainties to have any particular individual values, provided the product satisfies the restriction. Only measurements Q and R pass the stated compatibility test.

Explanation: \(\textbf{Use the equal-wavelength information:}\) For both particles, \[ \lambda=\frac hp \] Equal de Broglie wavelengths imply equal momentum magnitudes: \[ p_e=p_p \] In the non-relativistic approximation, \[ p=mv \] Therefore, \[ m_ev_e=m_pv_p \] Rearranging, \[ \frac{v_e}{v_p}=\frac{m_p}{m_e} \] \[ \frac{v_e}{v_p}=1836 \] \(\textbf{Calculate the electron's minimum momentum uncertainty:}\) Heisenberg's inequality gives \[ (\Delta p_{x,e})_{\min}=\frac{\hbar}{2\Delta x_e} \] With \(\Delta x_e=3d\), \[ (\Delta p_{x,e})_{\min}=\frac{\hbar}{6d} \] \(\textbf{Calculate the proton's minimum momentum uncertainty:}\) For \(\Delta x_p=d\), \[ (\Delta p_{x,p})_{\min}=\frac{\hbar}{2d} \] Therefore, \[ \frac{(\Delta p_{x,e})_{\min}}{(\Delta p_{x,p})_{\min}}=\frac13 \] \(\textbf{Convert the bounds into minimum velocity uncertainties:}\) For non-relativistic particles, \[ (\Delta v_x)_{\min}=\frac{(\Delta p_x)_{\min}}{m} \] Hence, \[ \frac{(\Delta v_{x,e})_{\min}}{(\Delta v_{x,p})_{\min}} =\frac{(\Delta p_{x,e})_{\min}}{(\Delta p_{x,p})_{\min}}\frac{m_p}{m_e} \] Substituting, \[ \frac{(\Delta v_{x,e})_{\min}}{(\Delta v_{x,p})_{\min}} =\frac13(1836) \] \[ \frac{(\Delta v_{x,e})_{\min}}{(\Delta v_{x,p})_{\min}}=612 \] \(\textbf{Distinguish speed from velocity uncertainty:}\) The actual speed comparison follows from the equal-momentum condition. The minimum uncertainty comparison follows from the separate position-localisation conditions. These comparisons need not produce the same numerical ratio. The minimum uncertainties are bounds, not claims that the actual experimental uncertainties necessarily attain equality. The example demonstrates how both particle mass and spatial localisation influence the uncertainty in velocity. \(\textbf{Final answer:}\) \[ \boxed{\frac{v_e}{v_p}=1836,\qquad\frac{(\Delta v_{x,e})_{\min}}{(\Delta v_{x,p})_{\min}}=612} \]

Explanation: \(\textbf{Determine the initial photoelectron kinetic energy:}\) Einstein's photoelectric equation gives \[ K_i=E_\gamma-\phi \] Substituting, \[ K_i=6.00-2.00 \] \[ K_i=4.00\,\mathrm{eV} \] \(\textbf{Translate the observed wavelength change:}\) For a non-relativistic electron, \[ \lambda=\frac{h}{\sqrt{2m_eK}} \] The final wavelength is half the initial wavelength: \[ \lambda_f=\frac{\lambda_i}{2} \] Therefore, \[ \frac{\lambda_f}{\lambda_i}=\sqrt{\frac{K_i}{K_f}} \] Substituting the wavelength ratio, \[ \frac12=\sqrt{\frac{K_i}{K_f}} \] Squaring, \[ \frac14=\frac{K_i}{K_f} \] \[ K_f=4K_i \] Hence, \[ K_f=4(4.00) \] \[ K_f=16.00\,\mathrm{eV} \] \(\textbf{Determine the additional energy supplied electrically:}\) The increase in electron kinetic energy is \[ \Delta K=K_f-K_i \] \[ \Delta K=16.00-4.00 \] \[ \Delta K=12.00\,\mathrm{eV} \] For an electron accelerated through potential difference \(V\), \[ \Delta K=eV \] Thus, an energy gain of \(12.00\,\mathrm{eV}\) corresponds to \[ V=12.00\,\mathrm{V} \] \(\textbf{Compare initial and final electron speeds:}\) For the same particle, \[ K=\frac12m_ev^2 \] Therefore, \[ \frac{v_f}{v_i}=\sqrt{\frac{K_f}{K_i}} \] \[ \frac{v_f}{v_i}=\sqrt{\frac{16.00}{4.00}} \] \[ \frac{v_f}{v_i}=2 \] \(\textbf{Interpret the two-stage energy process:}\) The photoelectric process supplies the electron's initial kinetic energy. The subsequent electric field supplies additional kinetic energy without creating another photoelectron. A twofold speed increase requires a fourfold kinetic-energy increase. Consequently, the additional electrical energy is three times the initial kinetic energy, not equal to it. The final wavelength change is consistent with the inverse relationship between wavelength and electron momentum. \(\textbf{Final answer:}\) \[ \boxed{V=12.0\,\mathrm{V},\qquad K_f=16.0\,\mathrm{eV},\qquad\frac{v_f}{v_i}=2} \]

Explanation: \(\textbf{Determine the initial minimum momentum uncertainty:}\) The position uncertainty is \[ \Delta x_i=0.100\,\mathrm{nm} \] Converting to SI units, \[ \Delta x_i=1.00\times10^{-10}\,\mathrm{m} \] The uncertainty inequality gives the lower bound \[ (\Delta p_x)_{\min}=\frac{\hbar}{2\Delta x} \] Substituting, \[ (\Delta p_{x,i})_{\min}=\frac{1.055\times10^{-34}}{2(1.00\times10^{-10})} \] \[ (\Delta p_{x,i})_{\min}=5.275\times10^{-25}\,\mathrm{kg\,m\,s^{-1}} \] \(\textbf{Calculate the initial kinetic-energy estimate:}\) Using the supplied kinetic-energy estimate, \[ K_{\min,i}=\frac{(\Delta p_{x,i})_{\min}^2}{2m_e} \] Substituting, \[ K_{\min,i}=\frac{(5.275\times10^{-25})^2}{2(9.11\times10^{-31})} \] \[ K_{\min,i}\approx1.527\times10^{-19}\,\mathrm{J} \] Converting to electronvolts, \[ K_{\min,i}=\frac{1.527\times10^{-19}}{1.602\times10^{-19}} \] \[ K_{\min,i}\approx0.953\,\mathrm{eV} \] \(\textbf{Apply the changed localisation condition:}\) The new uncertainty is \[ \Delta x_f=0.0500\,\mathrm{nm}=\frac{\Delta x_i}{2} \] The minimum momentum uncertainty is inversely proportional to \(\Delta x\). Therefore, \[ (\Delta p_{x,f})_{\min}=2(\Delta p_{x,i})_{\min} \] \(\textbf{Determine the new kinetic-energy estimate:}\) The estimated kinetic energy depends on the square of the momentum uncertainty. Hence, \[ \frac{K_{\min,f}}{K_{\min,i}}=\left(\frac{(\Delta p_{x,f})_{\min}}{(\Delta p_{x,i})_{\min}}\right)^2 \] \[ \frac{K_{\min,f}}{K_{\min,i}}=2^2 \] \[ K_{\min,f}=4(0.953) \] \[ K_{\min,f}\approx3.81\,\mathrm{eV} \] \(\textbf{Interpret the confinement effect:}\) Reducing position uncertainty increases the minimum spread in the corresponding momentum component. The kinetic-energy estimate rises quadratically because kinetic energy is proportional to momentum squared. This is a lower-bound estimate for the specified component, not a claim that every confined electron necessarily has exactly that kinetic energy. The calculation illustrates why increasingly precise localisation of a microscopic particle is associated with a significant momentum uncertainty. \(\textbf{Final answer:}\) \[ \boxed{K_{\min,i}\approx0.953\,\mathrm{eV},\qquad K_{\min,f}\approx3.81\,\mathrm{eV}} \]

Explanation: \(\textbf{Relate acceleration to the electron wavelength:}\) For electrons accelerated from rest through potential difference \(V\), \[ K=eV \] The non-relativistic de Broglie wavelength is \[ \lambda=\frac h{\sqrt{2m_e eV}} \] Therefore, \[ \lambda\propto\frac1{\sqrt V} \] \(\textbf{Analyse Change P:}\) The accelerating potential remains unchanged. The kinetic energy and momentum of individual electrons consequently remain unchanged. Their de Broglie wavelength is also unchanged. Under the stipulated diffraction geometry, the feature spacing therefore remains unchanged. The beam current is doubled, meaning that twice as much electron charge passes a reference point per unit time. For unchanged exposure time and detection efficiency, the detected electron count doubles. Thus, P increases the number of observed electron events without shifting the characteristic diffraction spacing. \(\textbf{Analyse Change Q:}\) The accelerating potential changes according to \[ V_Q=4V \] Therefore, \[ \frac{\lambda_Q}{\lambda_i}=\sqrt{\frac{V}{4V}} \] \[ \frac{\lambda_Q}{\lambda_i}=\frac12 \] The characteristic diffraction-feature spacing is assumed directly proportional to wavelength. Consequently, that spacing also becomes half its original value. The beam current remains unchanged, so the total detected electron count remains unchanged under the stated assumptions. \(\textbf{Interpret the distinct experimental controls:}\) Current controls the number of incident electrons per unit time. Accelerating potential controls their kinetic energy and consequently their de Broglie wavelength. A change in electron flux need not change the diffraction geometry. A change in electron wavelength can alter that geometry even without increasing the number of incident electrons. The result illustrates why wave behaviour and individual electron-detection events must be interpreted together rather than confused. \(\textbf{Final answer:}\) \[ \boxed{\mathrm P:\ N_{\mathrm{detected}}\rightarrow2N_{\mathrm{detected}},\ d\rightarrow d;\qquad\mathrm Q:\ N_{\mathrm{detected}}\rightarrow N_{\mathrm{detected}},\ d\rightarrow\frac d2} \]

Explanation: \(\textbf{Apply the equal-energy condition:}\) Each particle begins from rest and is accelerated through the same potential difference. Both have the same magnitude of charge. Therefore, each gains kinetic energy \[ K=|q|V \] Thus, \[ K_P=K_Q \] \(\textbf{Relate wavelength to particle mass:}\) For non-relativistic motion, \[ \lambda=\frac h{\sqrt{2mK}} \] At equal kinetic energy, \[ \lambda\propto\frac1{\sqrt m} \] Therefore, \[ \frac{\lambda_P}{\lambda_Q}=\sqrt{\frac{m_Q}{m_P}} \] Using the measured wavelengths, \[ \frac{0.200}{0.100}=\sqrt{\frac{m_Q}{m_P}} \] \[ 2=\sqrt{\frac{m_Q}{m_P}} \] Squaring, \[ \frac{m_Q}{m_P}=4 \] \(\textbf{Determine the final speed ratio:}\) The kinetic-energy relation is \[ K=\frac12mv^2 \] At equal kinetic energy, \[ v\propto\frac1{\sqrt m} \] Therefore, \[ \frac{v_Q}{v_P}=\sqrt{\frac{m_P}{m_Q}} \] \[ \frac{v_Q}{v_P}=\sqrt{\frac14} \] \[ \frac{v_Q}{v_P}=\frac12 \] \(\textbf{Check against the momentum ratio:}\) The de Broglie relation gives \[ p=\frac h\lambda \] Thus, \[ \frac{p_Q}{p_P}=\frac{\lambda_P}{\lambda_Q}=2 \] Using \(p=mv\), \[ \frac{p_Q}{p_P}=\frac{m_Q}{m_P}\frac{v_Q}{v_P} \] \[ \frac{p_Q}{p_P}=4\left(\frac12\right)=2 \] The independent momentum check agrees with the wavelength observation. \(\textbf{Interpret the result:}\) The heavier particle has the shorter de Broglie wavelength despite receiving the same kinetic energy. It also moves more slowly. Equal energy does not imply equal momentum or speed for particles of different masses. The measured wavelengths therefore permit inference of the mass ratio even without identifying the two particles chemically. \(\textbf{Final answer:}\) \[ \boxed{\frac{m_Q}{m_P}=4,\qquad\frac{v_Q}{v_P}=\frac12,\qquad\frac{K_Q}{K_P}=1} \]

Explanation: \(\textbf{Establish the starting uncertainty product:}\) The earlier measurement satisfies \[ \Delta x_i\,\Delta p_{x,i}=\frac{\hbar}{2} \] This is the lower bound permitted by the position-momentum uncertainty principle. \(\textbf{Translate the two claimed improvements:}\) The new position uncertainty is one-fourth of the original: \[ \Delta x_f=\frac{\Delta x_i}{4} \] The new momentum uncertainty is one-half of the original: \[ \Delta p_{x,f}=\frac{\Delta p_{x,i}}{2} \] \(\textbf{Calculate their combined product:}\) \[ \Delta x_f\,\Delta p_{x,f} =\frac{\Delta x_i}{4}\frac{\Delta p_{x,i}}{2} \] \[ \Delta x_f\,\Delta p_{x,f} =\frac18\Delta x_i\,\Delta p_{x,i} \] Using the original product, \[ \Delta x_f\,\Delta p_{x,f} =\frac18\left(\frac{\hbar}{2}\right) \] \[ \Delta x_f\,\Delta p_{x,f}=\frac{\hbar}{16} \] \(\textbf{Compare with the fundamental limit:}\) Heisenberg's inequality requires \[ \Delta x\,\Delta p_x\geq\frac{\hbar}{2} \] However, \[ \frac{\hbar}{16}\lt \frac{\hbar}{2} \] The proposed final uncertainties therefore violate the inequality. \(\textbf{Interpret instrumental improvement correctly:}\) Better experimental equipment may reduce additional uncertainties caused by imperfect calibration or limited resolution. It cannot produce a quantum state whose intrinsic position-momentum uncertainty product falls below the fundamental bound. Reducing one uncertainty from an already minimum-product state requires a compensating increase in the other. The uncertainty principle is not restricted to radiative transitions between atomic energy levels. It concerns the simultaneous specification of conjugate quantities in quantum mechanics. Thus, the student who distinguishes intrinsic quantum uncertainty from ordinary instrumental error is correct. \(\textbf{Final answer:}\) \[ \boxed{\Delta x_f\,\Delta p_{x,f}=\frac{\hbar}{16}\lt \frac{\hbar}{2}} \]

Explanation: \(\textbf{Apply the photoelectric threshold:}\) Einstein's photoelectric equation is \[ K=E_\gamma-\phi \] The work function is \[ \phi=2.00\,\mathrm{eV} \] Photoelectric emission requires photon energy at least equal to the work function. For an emitted electron having positive kinetic energy, \[ E_\gamma\gt 2.00\,\mathrm{eV} \] \(\textbf{Apply the diffraction requirement independently:}\) The apparatus requires \[ \lambda_e\geq0.120\,\mathrm{nm} \] The supplied electron relation is \[ \lambda_e=\frac{1.227}{\sqrt K}\,\mathrm{nm} \] Therefore, \[ \frac{1.227}{\sqrt K}\geq0.120 \] Rearranging, \[ \sqrt K\leq\frac{1.227}{0.120} \] \[ \sqrt K\leq10.225 \] Squaring, \[ K\leq(10.225)^2 \] \[ K\leq104.55\,\mathrm{eV} \] \(\textbf{Convert the kinetic-energy restriction into a photon-energy restriction:}\) Because \[ K=E_\gamma-2.00 \] the diffraction requirement becomes \[ E_\gamma-2.00\leq104.55 \] \[ E_\gamma\leq106.55\,\mathrm{eV} \] Rounding to the precision of the options, \[ E_\gamma\leq106.6\,\mathrm{eV} \] \(\textbf{Combine the two experimental conditions:}\) The photoelectric condition supplies the lower limit. The diffraction-resolution condition supplies the upper limit. The allowed range is therefore \[ 2.00\,\mathrm{eV}\lt E_\gamma\leq106.6\,\mathrm{eV} \] \(\textbf{Interpret the upper energy limit:}\) Increasing photon energy increases the kinetic energy of the emitted electron. That raises its momentum and decreases its de Broglie wavelength. Consequently, sufficiently energetic photoelectrons can satisfy the photoelectric condition but fail the particular diffraction apparatus's wavelength requirement. The upper limit comes from the stated resolving condition, not from any universal upper limit on photoelectric emission. The problem therefore requires two separate inequalities rather than a single threshold calculation. \(\textbf{Final answer:}\) \[ \boxed{2.00\,\mathrm{eV}\lt E_\gamma\leq106.6\,\mathrm{eV}} \]

Explanation: Statement I is correct because the observation of electron diffraction demonstrates wave-like behaviour. Diffraction is associated with the propagation and interference of waves. The observation cannot be explained adequately by treating electrons exclusively as classical point particles moving along independent definite trajectories. Statement II is correct because de Broglie proposed that moving material particles have an associated wavelength. For momentum magnitude \(p\), this wavelength is \(\lambda=h/p\). Therefore, changing the electron's momentum changes its associated wavelength. Statement III is incorrect because the uncertainty principle is not merely a limitation of inaccurate classical equations. It establishes a fundamental restriction on simultaneously specifying a position component and the corresponding momentum component. Even an ideal measurement arrangement cannot establish both quantities with arbitrarily small uncertainties. A precisely defined classical trajectory together with exactly specified momentum at every point is therefore incompatible with a complete quantum mechanical description of an atomic electron. Statement IV is correct because matter-wave behaviour and uncertainty are central reasons why the classical orbit picture is inadequate. An atomic orbital is associated with a wave function rather than an exactly measured circular path. The squared magnitude of the wave function provides the spatial probability density. This probabilistic interpretation does not mean that electronic states have arbitrary energies. Allowed atomic states can still possess quantised energies. The quantum mechanical model thus retains the successful idea of permitted electronic states while replacing the sharply defined trajectory assumption. The conclusions of the three experiments must be interpreted together: diffraction supports wave character, de Broglie's relation connects wavelength to momentum, and uncertainty restricts the classical position-momentum description. Only Statements I, II and IV are correct.

Explanation: \(\textbf{Infer the work functions from the first experiment:}\) Einstein's photoelectric equation is \[ K_{\max}=E_\gamma-\phi \] Therefore, \[ \phi=E_\gamma-K_{\max} \] For surface P, \[ \phi_P=5.00-2.80 \] \[ \phi_P=2.20\,\mathrm{eV} \] For surface Q, \[ \phi_Q=5.00-1.60 \] \[ \phi_Q=3.40\,\mathrm{eV} \] \(\textbf{Test surface P with the new photon energy:}\) The second experiment uses \[ E_{\gamma,2}=3.00\,\mathrm{eV} \] Because \[ 3.00\gt 2.20 \] the photons can eject electrons from P. Their maximum kinetic energy is \[ K_{\max,P}=3.00-2.20 \] \[ K_{\max,P}=0.80\,\mathrm{eV} \] \(\textbf{Test surface Q independently:}\) Surface Q requires at least \(3.40\,\mathrm{eV}\) per photon for photoelectric emission. The new photons carry only \(3.00\,\mathrm{eV}\). Thus, \[ E_{\gamma,2}\lt \phi_Q \] Photoelectric emission does not occur from Q under the stipulated one-photon conditions. It would be incorrect to report a negative photoelectron kinetic energy by mechanically subtracting the work function. \(\textbf{Interpret the changed conditions:}\) Both surfaces emitted electrons during the first experiment because the incident photon energy exceeded both work functions. Reducing photon energy changes their threshold eligibility. The metal with the smaller work function can continue emitting electrons after the other metal has fallen below threshold. Increasing the intensity of the second radiation would not enable photoelectric emission from Q while its individual photon energy remained insufficient. The comparison therefore requires determining each work function before applying the changed illumination. \(\textbf{Final answer:}\) \[ \phi_P=2.20\,\mathrm{eV},\quad\phi_Q=3.40\,\mathrm{eV},\quad K_{\max,P}=0.80\,\mathrm{eV} \]

Explanation: \(\textbf{Distinguish optical power from photon energy:}\) Optical power is energy delivered per unit time. If each photon has energy \(E_\gamma\), the number of incident photons per second is \[ \dot N=\frac{P_{\mathrm{optical}}}{E_\gamma} \] The two beams have equal optical powers but different photon energies. Therefore, they do not deliver equal numbers of photons per second. \(\textbf{Calculate the photon-rate ratio:}\) \[ \frac{\dot N_P}{\dot N_Q} =\frac{P_{\mathrm{optical}}/E_P}{P_{\mathrm{optical}}/E_Q} \] Cancelling the common power, \[ \frac{\dot N_P}{\dot N_Q}=\frac{E_Q}{E_P} \] Substituting, \[ \frac{\dot N_P}{\dot N_Q}=\frac{6.00}{3.00} \] \[ \frac{\dot N_P}{\dot N_Q}=2 \] \(\textbf{Convert photon rate into collected current:}\) Under the stated assumption, every photon ejects one collected electron. The collected current is \[ I=e\dot N \] Consequently, \[ \frac{I_P}{I_Q}=\frac{\dot N_P}{\dot N_Q} \] \[ \frac{I_P}{I_Q}=2 \] \(\textbf{Calculate the photoelectron kinetic energies:}\) Einstein's equation gives \[ K_{\max}=E_\gamma-\phi \] For P, \[ K_{\max,P}=3.00-2.00 \] \[ K_{\max,P}=1.00\,\mathrm{eV} \] For Q, \[ K_{\max,Q}=6.00-2.00 \] \[ K_{\max,Q}=4.00\,\mathrm{eV} \] Thus, \[ \frac{K_{\max,P}}{K_{\max,Q}}=\frac{1.00}{4.00} \] \[ \frac{K_{\max,P}}{K_{\max,Q}}=\frac14 \] \(\textbf{Interpret the contrasting outcomes:}\) Beam P produces more collected photoelectrons per second because its photons carry less energy at the same total power. Beam Q produces more energetic individual photoelectrons because its photons have greater energy above the same work function. The current ratio therefore cannot be inferred from the kinetic-energy ratio. Equal optical power does not mean equal photon flux, and equal photon flux would not by itself imply equal electron kinetic energies. The conclusion depends on the explicit collection and one-electron-per-photon assumptions. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_P}{I_Q}=2,\qquad\frac{K_{\max,P}}{K_{\max,Q}}=\frac14} \]

Explanation: \(\textbf{Express the linear photoelectric relationship:}\) Einstein's equation is \[ K_{\max}=h\nu-\phi \] At threshold, \[ \phi=h\nu_0 \] Therefore, \[ K_{\max}=h(\nu-\nu_0) \] The slope of the kinetic-energy-versus-frequency graph is Planck's constant. \(\textbf{Use the first experimental point:}\) At frequency \(f\), \[ hf-\phi=eV \] \(\textbf{Use the second experimental point:}\) At frequency \(2f\), \[ 2hf-\phi=3eV \] Subtracting the first equation from the second, \[ hf=2eV \] \(\textbf{Determine the work function:}\) Substitute the value of \(hf\) into the first equation: \[ 2eV-\phi=eV \] \[ \phi=eV \] \(\textbf{Identify the threshold frequency:}\) The threshold condition gives \[ h\nu_0=\phi \] Therefore, \[ \nu_0=\frac{eV}{h} \] Since \(hf=2eV\), \[ \nu_0=\frac f2 \] \(\textbf{Predict the third experimental result:}\) The new incident frequency is \[ \nu_3=\frac{3f}{2} \] Its maximum photoelectron kinetic energy is \[ K_{\max,3}=h\left(\frac{3f}{2}\right)-\phi \] Substituting \(hf=2eV\) and \(\phi=eV\), \[ K_{\max,3}=\frac32(2eV)-eV \] \[ K_{\max,3}=2eV \] \(\textbf{Check the physical threshold:}\) The third frequency is greater than the inferred threshold \(f/2\). Photoelectric emission is therefore permitted. The predicted kinetic energy lies between those measured at \(f\) and \(2f\), consistent with the linear graph. The work function remains constant because the metal has not changed. The problem requires using two observations to infer a property of the surface before predicting a changed-frequency result. \(\textbf{Final answer:}\) \[ \boxed{\nu_0=\frac f2,\qquad K_{\max,3}=2eV} \]

Explanation: Statement I is correct because the frequency of monochromatic electromagnetic radiation remains unchanged when it passes between stationary transparent media. The propagation speed and wavelength can change, but the number of wave oscillations per second remains the same. Statement II is incorrect because photon energy is determined by \[ E_\gamma=h\nu \] Since frequency remains unchanged, the energy of each photon remains unchanged under the stated assumptions. The relation \(E=hc/\lambda\) uses the vacuum speed of light together with the corresponding vacuum wavelength. It must not be applied using the vacuum speed \(c\) and an altered wavelength in a material medium. Statement III is correct because the photoelectric threshold depends on the photon energy and the metal's work function. Both quantities are stipulated to remain unchanged. Consequently, a beam that initially exceeds the threshold continues to exceed it after the change of medium. Statement IV is correct because the medium wavelength is \[ \lambda_m=\frac34\lambda_a \] The wavenumber in each medium is the reciprocal of its local wavelength. Therefore, \[ \frac{\tilde\nu_m}{\tilde\nu_a} =\frac{\lambda_a}{\lambda_m} \] \[ \frac{\tilde\nu_m}{\tilde\nu_a}=\frac43 \] The change in wavelength is accompanied by the corresponding change in propagation speed. It is not evidence of a change in photon energy. This distinction becomes especially important when electromagnetic wave relations are combined with the photoelectric effect. Only Statements I, III and IV are correct.

Explanation: \(\textbf{Apply the threshold-wavelength condition:}\) The threshold wavelength is \[ \lambda_0=500\,\mathrm{nm} \] Photoelectric emission requires photon energy at least equal to the work function: \[ \frac{hc}{\lambda}\geq\frac{hc}{\lambda_0} \] Therefore, \[ \lambda\leq\lambda_0 \] Source P has wavelength \(400\,\mathrm{nm}\), which is below the threshold wavelength. Its photons are energetic enough to eject electrons. Source Q has wavelength \(600\,\mathrm{nm}\), which exceeds the threshold wavelength. Its photons are below the required energy threshold. Thus, only P can cause photoelectric emission. \(\textbf{Determine photon arrival rate from optical power:}\) For monochromatic radiation, \[ P_{\mathrm{optical}}=\dot N E_\gamma \] With \[ E_\gamma=\frac{hc}{\lambda} \] the photon arrival rate becomes \[ \dot N=\frac{P_{\mathrm{optical}}\lambda}{hc} \] \(\textbf{Compare the two arrival rates:}\) \[ \frac{\dot N_Q}{\dot N_P} =\frac{P_Q\lambda_Q}{P_P\lambda_P} \] Substituting, \[ \frac{\dot N_Q}{\dot N_P} =\frac{16.0(600)}{8.0(400)} \] \[ \frac{\dot N_Q}{\dot N_P}=3 \] \(\textbf{Interpret the two experimental outcomes:}\) Source Q delivers three times as many photons per second as P. Nevertheless, Q produces no photoelectric emission because its individual photons lack the required energy. The greater photon arrival rate cannot compensate for this failure under the stipulated one-photon conditions. Source P has lower optical power but a shorter wavelength and therefore greater photon energy. The result separates the number of incident photons from the capacity of an individual photon to eject an electron. Optical power alone does not determine whether a metal emits photoelectrons. \(\textbf{Final answer:}\) \[ \dot N_Q/\dot N_P=3,\qquad\lambda_P\lt \lambda_0\lt \lambda_Q \]

Explanation: \(\textbf{Express the work functions through threshold frequency:}\) For X, \[ \phi_X=h\nu_X \] For Y, \[ \phi_Y=h\nu_Y \] Using the given relationship, \[ \phi_Y=\frac32h\nu_X \] \(\textbf{Compare the kinetic energies under the initial illumination:}\) The incident frequency is \(2\nu_X\). For X, \[ K_X=h(2\nu_X)-h\nu_X \] \[ K_X=h\nu_X \] For Y, \[ K_Y=h(2\nu_X)-\frac32h\nu_X \] \[ K_Y=\frac12h\nu_X \] Therefore, \[ \frac{K_X}{K_Y}=\frac{h\nu_X}{h\nu_X/2} \] \[ \frac{K_X}{K_Y}=2 \] Both surfaces emit electrons in the first experiment because \(2\nu_X\) exceeds both threshold frequencies. \(\textbf{Apply the changed incident frequency:}\) The new frequency is \[ \nu_{\mathrm{new}}=\frac54\nu_X \] For X, \[ \frac54\nu_X\gt \nu_X \] Hence, X remains above threshold. Its new maximum photoelectron kinetic energy is \[ K_{X,\mathrm{new}}=h\left(\frac54\nu_X-\nu_X\right) \] \[ K_{X,\mathrm{new}}=\frac14h\nu_X \] For Y, the threshold is \(3\nu_X/2\). Since \[ \frac54\nu_X\lt \frac32\nu_X \] the new radiation falls below Y's photoelectric threshold. Y therefore emits no photoelectrons under the stipulated one-photon conditions. \(\textbf{Interpret the comparison:}\) A common decrease in incident frequency does not necessarily produce the same qualitative response from different metals. A surface with a smaller work function can continue emitting after another surface has fallen below threshold. The comparison requires checking the threshold separately for each metal after calculating the initial kinetic-energy ratio. The photon frequency, rather than the beam intensity alone, determines whether the photoelectric process remains energetically permitted. \(\textbf{Final answer:}\) \[ \boxed{K_X:K_Y=2:1,\qquad K_{X,\mathrm{new}}=\frac14h\nu_X} \]

Explanation: \(\textbf{Check threshold eligibility before calculating currents:}\) The work function is \[ \phi=2.50\,\mathrm{eV} \] A photon can eject an electron only if its energy meets or exceeds the work function. For beam P, \[ E_P=2.00\,\mathrm{eV}\lt \phi \] Therefore, P produces no photoelectric emission despite having the largest photon arrival rate. Beam Q has energy \(3.00\,\mathrm{eV}\), which exceeds the threshold. Beam R has energy \(5.00\,\mathrm{eV}\), which also exceeds the threshold. \(\textbf{Convert eligible photon arrival rates into current:}\) With the stated one-electron-per-photon collection assumption, \[ I=e\dot N \] Therefore, \[ I_P=0 \] \[ I_Q=e(4N)=4eN \] \[ I_R=e(2N)=2eN \] The current ratio is \[ I_P:I_Q:I_R=0:4:2 \] \[ I_P:I_Q:I_R=0:2:1 \] \(\textbf{Determine the maximum photoelectron kinetic energies:}\) Einstein's photoelectric equation is \[ K_{\max}=E_\gamma-\phi \] For Q, \[ K_{\max,Q}=3.00-2.50 \] \[ K_{\max,Q}=0.50\,\mathrm{eV} \] For R, \[ K_{\max,R}=5.00-2.50 \] \[ K_{\max,R}=2.50\,\mathrm{eV} \] \(\textbf{Interpret the different rankings:}\) Beam P has the greatest photon arrival rate but produces no electrons because each photon lies below threshold. Beam Q produces more photoelectrons per second than R because it delivers more eligible photons. However, the individual photoelectrons produced by R have greater maximum kinetic energy. Thus, the ranking of photon arrival rates, collected currents and photoelectron kinetic energies need not be the same. This distinction is central to the quantum explanation of the photoelectric effect. The threshold check must precede the photon-count and kinetic-energy calculations. \(\textbf{Final answer:}\) \[ \boxed{I_P:I_Q:I_R=0:2:1,\qquad K_{\max,Q}=0.50\,\mathrm{eV},\quad K_{\max,R}=2.50\,\mathrm{eV}} \]

Explanation: \(\textbf{Infer the allowed work-function intervals:}\) Photoelectric emission is possible when the incident photon energy is at least equal to the work function. For surface A, emission does not occur at \(2.0\,\mathrm{eV}\). Therefore, its work function exceeds that energy under the stated ideal observations. Emission occurs at \(3.0\,\mathrm{eV}\), so A's work function cannot exceed \(3.0\,\mathrm{eV}\). Thus, \[ 2.0\,\mathrm{eV}\lt \phi_A\leq3.0\,\mathrm{eV} \] For surface B, emission does not occur at \(3.0\,\mathrm{eV}\), but occurs at \(4.0\,\mathrm{eV}\). Consequently, \[ 3.0\,\mathrm{eV}\lt \phi_B\leq4.0\,\mathrm{eV} \] \(\textbf{Test the fourth photon energy against these intervals:}\) The proposed new photon energy is \[ E_{\gamma,4}=3.5\,\mathrm{eV} \] Every work function permitted for A is below \(3.5\,\mathrm{eV}\). Therefore, A must exhibit photoelectric emission. For B, however, the permitted interval extends both below and above \(3.5\,\mathrm{eV}\). If \(\phi_B\leq3.5\,\mathrm{eV}\), B can emit. If \(\phi_B\gt 3.5\,\mathrm{eV}\), B cannot emit under the stipulated one-photon conditions. The original observations do not distinguish between these possibilities. \(\textbf{Interpret the experimental inference:}\) A finite set of threshold observations can establish bounds on a metal's work function without determining its exact value. A failed emission experiment does not mean that all photon energies slightly above the tested energy must also fail. Similarly, an emitting surface does not reveal an exact threshold unless additional information is supplied. The fourth experiment therefore has a certain outcome for A but an undetermined outcome for B. The distinction comes from incomplete threshold information, not from any uncertainty about Einstein's photoelectric equation. This is why only option D is necessarily supported.