Dual Nature of Matter and Radiation MCQs with Answers – Part 5 (NEET Physics)

Explanation: \(\textbf{Determine the work function from the spectral cutoff:}\) The threshold wavelength is: \[ \lambda_0=500\,\text{nm} \] At threshold: \[ \phi=\frac{hc}{\lambda_0} \] Substituting: \[ \phi=\frac{1240\,\text{eV nm}}{500\,\text{nm}} \] \[ \phi=2.48\,\text{eV} \] \(\textbf{Determine photon energy at the measured graph point:}\) For: \[ \lambda=400\,\text{nm} \] the incident photon energy is: \[ E_\gamma=\frac{hc}{\lambda} \] \[ E_\gamma=\frac{1240}{400}\,\text{eV} \] \[ E_\gamma=3.10\,\text{eV} \] In joules: \[ E_\gamma=3.10(1.60\times10^{-19}) \] \[ E_\gamma=4.96\times10^{-19}\,\text{J} \] \(\textbf{Find the photon arrival rate at fixed optical power:}\) \[ R_\gamma=\frac{P}{E_\gamma} \] Substituting: \[ R_\gamma=\frac{0.620\times10^{-3}}{4.96\times10^{-19}} \] \[ R_\gamma=1.25\times10^{15}\,\text{s}^{-1} \] \(\textbf{Recover the photoelectron emission efficiency:}\) At saturation: \[ I_s=e\eta R_\gamma \] Therefore: \[ \eta=\frac{I_s}{eR_\gamma} \] Substituting: \[ \eta=\frac{100\times10^{-6}}{(1.60\times10^{-19})(1.25\times10^{15})} \] \[ \eta=\frac{100}{200} \] \[ \eta=0.50 \] \[ \eta=50\% \] \(\textbf{Predict current at the new wavelength:}\) At fixed optical power and efficiency: \[ I_s=\frac{e\eta P}{hc/\lambda} \] Thus: \[ I_s\propto\lambda \] within the above-threshold range. For: \[ \lambda'=450\,\text{nm} \] we obtain: \[ I'_s=100\,\mu\text{A}\left(\frac{450}{400}\right) \] \[ I'_s=112.5\,\mu\text{A} \] \(\textbf{Interpret the end of the straight-line branch:}\) The linear above-threshold relationship cannot be extended physically beyond the threshold wavelength. Longer-wavelength photons individually carry insufficient energy to overcome the work function, even though a longer wavelength would imply more photons per second at the same optical power. The current therefore disappears rather than following the extrapolated straight line. \(\textbf{Final answer:}\) \[ \boxed{\phi=2.48\,\text{eV},\qquad\eta=50\%,\qquad I_s(450\,\text{nm})=112.5\,\mu\text{A}} \]

Explanation: \(\textbf{Identify the battery voltage available to the tube:}\) The battery emf is: \[ \mathcal E=3.0\,\text{V} \] At a steady photocurrent \(I\), the resistor has voltage drop: \[ V_R=IR \] Therefore, the actual collector-emitter voltage is: \[ V_C=\mathcal E-IR \] \(\textbf{Apply the specified saturation condition:}\) Complete collection requires: \[ V_C\geq+1.0\,\text{V} \] For the limiting resistance, the tube just reaches saturation at: \[ I=I_s=100\,\mu\text{A} \] and: \[ V_C=1.0\,\text{V} \] Thus: \[ \mathcal E-I_sR_{\max}=1.0\,\text{V} \] \(\textbf{Determine the greatest permissible resistance:}\) Rearranging: \[ R_{\max}=\frac{\mathcal E-1.0\,\text{V}}{I_s} \] Substituting: \[ R_{\max}=\frac{3.0-1.0}{100\times10^{-6}}\,\Omega \] \[ R_{\max}=\frac{2.0}{1.00\times10^{-4}}\,\Omega \] \[ R_{\max}=2.0\times10^4\,\Omega \] \[ R_{\max}=20\,\text{k}\Omega \] \(\textbf{Determine the intrinsic stopping potential independently:}\) Einstein's photoelectric equation gives: \[ K_{\max}=E_\gamma-\phi \] Substituting: \[ K_{\max}=3.0-2.0 \] \[ K_{\max}=1.0\,\text{eV} \] Therefore: \[ V_0=1.0\,\text{V} \] \(\textbf{Check the effect of a larger series resistance:}\) If the resistor were greater than \(20\,\text{k}\Omega\), the assumed \(100\,\mu\text{A}\) current would cause a voltage drop exceeding \(2.0\,\text{V}\). The tube would then receive less than the specified \(+1.0\,\text{V}\) required for saturation. Therefore, the full saturation current could not be sustained under the stated current-voltage characteristic. \(\textbf{Physical interpretation:}\) The external resistor reduces the collecting voltage available at a given current. It can prevent complete electron collection without changing the emission energy supplied by individual photons. The resistance limit is determined electrically, whereas the photoelectric stopping potential is determined by photon energy and work function. \(\textbf{Final answer:}\) \[ \boxed{R_{\max}=20\,\text{k}\Omega,\qquad V_0=1.0\,\text{V}} \]

Explanation: Ordinary photoelectric emission requires an incident photon to carry at least the work-function energy needed for electron escape. In Experiment I, the visible photon energy is smaller than P's work function, so that radiation cannot produce ordinary single-photon photoelectric emission from P. In Experiment II, the ultraviolet photons exceed P's work function and can eject electrons. Because P is isolated, the escape of negatively charged electrons leaves the metal with positive net charge. In Experiment III, the visible photon energy exceeds Q's smaller work function. Visible light can therefore produce photoelectric emission from Q, showing that ultraviolet illumination is not universally necessary for every metal. Experiment IV also permits emission because ultraviolet photon energy is greater than Q's work function. Thus, the plates can initially acquire positive charge in Experiments II, III and IV. The result is consistent with the charge-loss observations associated with Hallwachs and with material-dependent photoelectric thresholds. It also explains why alkali metals can be sensitive to visible light while other metallic surfaces may require ultraviolet radiation. Photons do not deliver positive electric charge to the isolated plates; the positive charging results from electrons leaving. As charging proceeds, the developing electrostatic potential can eventually inhibit further successful electron escape, but that later effect is excluded from the initial comparison.

Explanation: \(\textbf{Determine the electrostatic stopping condition:}\) Einstein's photoelectric equation gives: \[ K_{\max}=E_\gamma-\phi \] Substituting: \[ K_{\max}=3.0-2.0 \] \[ K_{\max}=1.0\,\text{eV} \] The isolated emitter stops transferring electrons to the grounded collector when its positive potential reaches: \[ V_{\max}=1.0\,\text{V} \] \(\textbf{Calculate the corresponding charge limit:}\) The capacitance is: \[ C=2.0\,\text{pF} \] Using: \[ Q_{\max}=CV_{\max} \] we obtain: \[ Q_{\max}=(2.0\times10^{-12})(1.0)\,\text{C} \] \[ Q_{\max}=2.0\times10^{-12}\,\text{C} \] \(\textbf{Calculate the unrestricted photoelectric current:}\) The incident photon energy in joules is: \[ E_\gamma=3.0(1.60\times10^{-19}) \] \[ E_\gamma=4.80\times10^{-19}\,\text{J} \] The photon arrival rate is: \[ R_\gamma=\frac{P}{E_\gamma} \] Substituting: \[ R_\gamma=\frac{0.30\times10^{-3}}{4.80\times10^{-19}} \] \[ R_\gamma=6.25\times10^{14}\,\text{s}^{-1} \] The emission rate under the stated ideal model is: \[ R_e=\eta R_\gamma \] \[ R_e=0.10(6.25\times10^{14}) \] \[ R_e=6.25\times10^{13}\,\text{s}^{-1} \] Therefore, the transfer-current magnitude is: \[ I=eR_e \] \[ I=(1.60\times10^{-19})(6.25\times10^{13}) \] \[ I=1.00\times10^{-5}\,\text{A} \] \[ I=10.0\,\mu\text{A} \] \(\textbf{Determine the charging time:}\) Under the stated constant-rate approximation: \[ Q_{\max}=It \] Therefore: \[ t=\frac{Q_{\max}}{I} \] Substituting: \[ t=\frac{2.0\times10^{-12}}{1.00\times10^{-5}}\,\text{s} \] \[ t=2.00\times10^{-7}\,\text{s} \] \[ t=200\,\text{ns} \] \(\textbf{Apply the optical-power change:}\) Doubling power at unchanged photon energy and efficiency doubles the photoelectron-transfer rate: \[ I'=2I \] The electrostatic charge limit does not change: \[ Q'_{\max}=Q_{\max} \] Thus: \[ t'=\frac{Q_{\max}}{2I} \] \[ t'=\frac t2=100\,\text{ns} \] The final stopping potential remains: \[ V'_{\max}=1.0\,\text{V} \] \(\textbf{Physical interpretation:}\) The charging rate depends on incident photon number per second, but the final electrostatic potential depends on the maximum initial energy of an emitted photoelectron. Increasing intensity therefore changes how quickly the isolated system reaches its stopping condition without changing the stopping potential itself. \(\textbf{Final answer:}\) \[ \boxed{t=200\,\text{ns},\qquad t'=100\,\text{ns},\qquad V'_{\max}=1.0\,\text{V}} \]

Explanation: The incident ultraviolet photons supply energy to electrons in the illuminated emitter. When the photon energy exceeds the work function, some electrons can escape from the metallic surface. The collector is maintained at a higher electric potential than the emitter, so the electric field assists negatively charged photoelectrons travelling towards the collector. Electron motion through the evacuated gap is therefore from emitter to collector. Conventional current is defined in the direction of positive-charge transport, so its direction through the gap is opposite to the electron motion. In the connected arrangement, the external battery and circuit permit charge redistribution that maintains steady current and prevents unrestricted charge accumulation on the electrodes. When the collector is disconnected, arriving electrons deposit negative charge on it. The isolated collector therefore initially becomes more negatively charged. This negative charging can eventually create a retarding potential that inhibits further collection, depending on the electron energies and capacitance. The emitter is the source of emitted electrons, not the collector, under the stated illumination conditions. Nor do photons themselves transport negative electric charge across the gap. The experiment distinguishes photon energy transfer, electron motion, conventional current direction and electrostatic charging. These are separate but connected parts of the photoelectric process.

Explanation: \(\textbf{Calculate the absorbed photon arrival rate:}\) The incident optical power is: \[ P=1.20\times10^{-3}\,\text{W} \] The photon energy is: \[ E_\gamma=4.0\,\text{eV} \] In joules: \[ E_\gamma=4.0(1.60\times10^{-19}) \] \[ E_\gamma=6.40\times10^{-19}\,\text{J} \] Therefore: \[ R_\gamma=\frac{P}{E_\gamma} \] \[ R_\gamma=\frac{1.20\times10^{-3}}{6.40\times10^{-19}} \] \[ R_\gamma=1.875\times10^{15}\,\text{s}^{-1} \] \(\textbf{Determine photoelectron emission rate:}\) The quantum efficiency is: \[ \eta=0.25 \] Thus: \[ R_e=\eta R_\gamma \] \[ R_e=0.25(1.875\times10^{15}) \] \[ R_e=4.6875\times10^{14}\,\text{s}^{-1} \] \(\textbf{Calculate the saturation photocurrent:}\) At complete collection: \[ I_s=eR_e \] Substituting: \[ I_s=(1.60\times10^{-19})(4.6875\times10^{14}) \] \[ I_s=7.50\times10^{-5}\,\text{A} \] \[ I_s=75\,\mu\text{A} \] \(\textbf{Determine kinetic energy carried by each emitted electron:}\) Einstein's equation gives: \[ K_{\max}=E_\gamma-\phi \] \[ K_{\max}=4.0-2.0 \] \[ K_{\max}=2.0\,\text{eV} \] Converting: \[ K_{\max}=3.20\times10^{-19}\,\text{J} \] \(\textbf{Calculate the total emitted-electron kinetic-energy power:}\) \[ P_{K,e}=R_eK_{\max} \] Substituting: \[ P_{K,e}=(4.6875\times10^{14})(3.20\times10^{-19}) \] \[ P_{K,e}=1.50\times10^{-4}\,\text{W} \] \[ P_{K,e}=0.15\,\text{mW} \] \(\textbf{Apply energy conservation to the metal:}\) All incident optical power is absorbed. The only specified energy leaving the metal is the translational kinetic energy of emitted electrons. The work-function energy required to free electrons is not an additional energy output; it is part of the energy transferred into the electron-metal system during emission. Therefore, the net power remaining in the metal is: \[ P_{\mathrm{remain}}=P-P_{K,e} \] \[ P_{\mathrm{remain}}=1.20-0.15 \] \[ P_{\mathrm{remain}}=1.05\,\text{mW} \] The retained fraction is: \[ f=\frac{P_{\mathrm{remain}}}{P} \] \[ f=\frac{1.05}{1.20} \] \[ f=0.875=87.5\% \] \(\textbf{Physical interpretation:}\) Only a fraction of absorbed photons produces escaping electrons. Even those successful electrons carry away only the excess photon energy beyond the work function. The total outgoing kinetic-energy power can therefore be much smaller than the incident optical power. The remaining energy is retained within the illuminated material under the specified idealised energy-accounting assumptions. \(\textbf{Final answer:}\) \[ \boxed{I_s=75\,\mu\text{A},\qquad P_{K,e}=0.15\,\text{mW},\qquad f_{\mathrm{remain}}=87.5\%} \]

Explanation: \(\textbf{Determine the original absorbed-photon fraction:}\) Let the number of photons initially reaching the film per second be \(R\). The single-pass absorption probability is: \[ a=\frac14 \] Without the mirror: \[ R_{\mathrm{abs},1}=aR \] \[ R_{\mathrm{abs},1}=\frac R4 \] \(\textbf{Find the photons surviving the first traversal:}\) The transmission probability is: \[ 1-a=\frac34 \] Therefore, the photon rate reaching the mirror is: \[ R_{\mathrm{trans}}=\frac{3R}{4} \] The perfect mirror returns these photons to the film for one further traversal. \(\textbf{Determine additional absorption on return:}\) Each returning photon again has absorption probability \(1/4\). Thus: \[ R_{\mathrm{abs},2}=\frac14R_{\mathrm{trans}} \] Substituting: \[ R_{\mathrm{abs},2}=\frac14\left(\frac{3R}{4}\right) \] \[ R_{\mathrm{abs},2}=\frac{3R}{16} \] \(\textbf{Combine the two absorption opportunities:}\) The total absorbed rate with the mirror is: \[ R_{\mathrm{abs,mirror}}=\frac R4+\frac{3R}{16} \] \[ R_{\mathrm{abs,mirror}}=\frac{4R+3R}{16} \] \[ R_{\mathrm{abs,mirror}}=\frac{7R}{16} \] \(\textbf{Compare emitted-electron rates and saturation currents:}\) The emission probability per absorbed photon is unchanged. Therefore: \[ I_s=e\eta R_{\mathrm{abs}} \] Hence: \[ \frac{I_{s,\mathrm{mirror}}}{I_{s,\mathrm{original}}}=\frac{7R/16}{R/4} \] \[ \frac{I_{s,\mathrm{mirror}}}{I_{s,\mathrm{original}}}=\frac74 \] \(\textbf{Check the maximum photoelectron energy:}\) The mirror returns photons at unchanged frequency. Therefore, the energy of each absorbed photon remains: \[ E_\gamma=h\nu \] With unchanged work function: \[ K_{\max}=h\nu-\phi \] Thus, the stopping-potential magnitude remains unchanged. \(\textbf{Physical interpretation:}\) The mirror increases the chance that an incident photon will be absorbed by supplying a second opportunity for absorption. It does not increase the energy carried by an individual photon. More photoelectrons can therefore be emitted per second without increasing their maximum kinetic energy. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_{s,\mathrm{mirror}}}{I_{s,\mathrm{original}}}=\frac74,\qquad V_{0,\mathrm{mirror}}=V_{0,\mathrm{original}}} \]

Explanation: The electroscope observation indicates that the illuminated plate becomes less negatively charged when suitable ultraviolet radiation strikes it. This is consistent with electrons escaping from the metal. Electrons possess negative charge, so their departure increases the algebraic charge of the remaining isolated plate. Therefore, conclusions P and Q are supported. In the photoelectric tube, emitted electrons can travel through the evacuated gap towards a suitably maintained collector. The external circuit permits continuing charge redistribution, allowing a steady photocurrent to be measured while suitable illumination continues. Conclusion R is therefore also supported. However, observing electrical current does not imply that incident photons themselves transport electric charge through the apparatus. Photons are electrically neutral. Their role is to supply energy to electrons during the photoelectric interaction. The electrons are the charged carriers travelling between electrodes. Conclusion S therefore confuses energy transfer during emission with subsequent charge transport. Comparing an isolated electroscope plate with a connected photoelectric tube reveals the same microscopic emission process through two different electrical measurements. The supported sequence is P, Q and R, while S is unsupported.

Explanation: \(\textbf{Identify the two coupled electrical relationships:}\) The modelled collected-current relation is: \[ I=I_s\left(1-\frac{V_r}{V_0}\right) \] The feedback circuit imposes: \[ V_r=IR \] The photocurrent and retarding voltage therefore cannot be calculated independently. \(\textbf{Substitute the feedback voltage:}\) Replacing \(V_r\): \[ I=I_s\left(1-\frac{IR}{V_0}\right) \] Expanding: \[ I=I_s-\frac{I_sIR}{V_0} \] Collecting the terms containing \(I\): \[ I\left(1+\frac{I_sR}{V_0}\right)=I_s \] Therefore: \[ I=\frac{I_s}{1+I_sR/V_0} \] \(\textbf{Evaluate the feedback strength:}\) The saturation current is: \[ I_s=20\times10^{-6}\,\text{A} \] The resistance is: \[ R=50\times10^3\,\Omega \] Thus: \[ I_sR=(20\times10^{-6})(50\times10^3)\,\text{V} \] \[ I_sR=1.0\,\text{V} \] Therefore: \[ \frac{I_sR}{V_0}=\frac{1.0}{2.0} \] \[ \frac{I_sR}{V_0}=0.50 \] \(\textbf{Determine the steady photocurrent:}\) \[ I=\frac{20\,\mu\text{A}}{1+0.50} \] \[ I=\frac{20}{1.5}\,\mu\text{A} \] \[ I\approx13.33\,\mu\text{A} \] \(\textbf{Recover the self-consistent retarding voltage:}\) Using: \[ V_r=IR \] we obtain: \[ V_r=(13.33\times10^{-6})(50\times10^3)\,\text{V} \] \[ V_r\approx0.667\,\text{V} \] \(\textbf{Check against the photoelectric current relation:}\) \[ I=20\left(1-\frac{0.667}{2.0}\right)\,\mu\text{A} \] \[ I\approx13.33\,\mu\text{A} \] The two circuit relationships agree. \(\textbf{Physical interpretation:}\) A larger collected current produces a larger retarding voltage through the feedback resistor. That voltage reduces the fraction of emitted electrons reaching the collector, which in turn limits the current. The steady state is therefore determined by simultaneous electrical and photoelectric conditions rather than by the saturation current alone. \(\textbf{Final answer:}\) \[ \boxed{I\approx13.3\,\mu\text{A},\qquad V_r\approx0.667\,\text{V}} \]

Explanation: \(\textbf{Determine the emitted electron's maximum kinetic energy:}\) Einstein's equation gives: \[ K_{\max}=E_\gamma-\phi \] Substituting: \[ K_{\max}=4.0-2.0 \] \[ K_{\max}=2.0\,\text{eV} \] Converting into joules: \[ K_{\max}=2.0(1.60\times10^{-19}) \] \[ K_{\max}=3.20\times10^{-19}\,\text{J} \] \(\textbf{Calculate the electron momentum:}\) For a non-relativistic electron: \[ p_e=\sqrt{2m_eK_{\max}} \] Substituting: \[ p_e=\sqrt{2(9.10\times10^{-31})(3.20\times10^{-19})} \] \[ p_e\approx7.63\times10^{-25}\,\text{kg m s}^{-1} \] \(\textbf{Determine incident photon momentum:}\) For a photon: \[ p_\gamma=\frac{E_\gamma}{c} \] The photon energy in joules is: \[ E_\gamma=4.0(1.60\times10^{-19}) \] \[ E_\gamma=6.40\times10^{-19}\,\text{J} \] Therefore: \[ p_\gamma=\frac{6.40\times10^{-19}}{3.0\times10^8} \] \[ p_\gamma\approx2.13\times10^{-27}\,\text{kg m s}^{-1} \] \(\textbf{Compare the two momentum magnitudes:}\) \[ \frac{p_e}{p_\gamma}=\frac{7.63\times10^{-25}}{2.13\times10^{-27}} \] \[ \frac{p_e}{p_\gamma}\approx358 \] Thus, the electron momentum is approximately \(360\) times the photon momentum. \(\textbf{Apply momentum conservation to the metal recoil:}\) Choose the incident photon direction as positive. Momentum conservation gives: \[ p_\gamma=p_e+p_M \] Therefore: \[ p_M=p_\gamma-p_e \] Since: \[ p_e\gt p_\gamma \] the metal recoil momentum is negative: \[ p_M\lt 0 \] Thus, the metal recoils opposite the photon direction. \(\textbf{Physical interpretation:}\) A photoelectron can emerge with momentum much greater than the incident photon momentum because the emitting metal participates in momentum conservation. The work function and electron kinetic energy are determined through energy transfer, while the recoiling metal carries the momentum required to balance the process. The metal's large mass allows significant recoil momentum to be associated with negligibly small recoil kinetic energy. \(\textbf{Final answer:}\) \[ \frac{p_e}{p_\gamma}\approx360,\qquad\text{metal recoil momentum is opposite to incident light} \]

Explanation: \(\textbf{Relate magnetic curvature to electron momentum:}\) For motion perpendicular to a uniform magnetic field: \[ evB=\frac{m_ev^2}{r} \] Thus: \[ r=\frac{m_ev}{eB} \] Since: \[ p_e=m_ev \] the radius is: \[ r=\frac{p_e}{eB} \] \(\textbf{Identify the effect of magnetic-field strength:}\) The magnetic force changes electron velocity direction but does no work. Therefore: \[ K'=K \] and: \[ p'_e=p_e \] At unchanged electron momentum: \[ r\propto\frac1B \] \(\textbf{Apply the original trajectory geometry:}\) Initially: \[ r_1=1.5d \] Since: \[ r_1\gt d \] the circular trajectory can reach the collector plane \(x=d\). The stipulated transverse extent of the collector allows these electrons to be collected. \(\textbf{Apply the doubled magnetic field:}\) The new field is: \[ B_2=2B_1 \] Therefore: \[ r_2=\frac{r_1}{2} \] Substituting: \[ r_2=\frac{1.5d}{2} \] \[ r_2=0.75d \] \(\textbf{Test whether the new orbit can reach the collector:}\) An electron launched along positive \(x\) into a perpendicular magnetic field follows a circle whose maximum forward displacement from its starting point is its orbit radius. Therefore: \[ x_{\max}=r_2 \] Since: \[ r_2=0.75d\lt d \] the electron cannot reach the collector plane. The collected photocurrent can therefore fall to zero under the stipulated ideal conditions. \(\textbf{Check the intrinsic photoelectric energy:}\) The magnetic field does no work. Thus, the maximum initial photoelectron kinetic energy remains unchanged. Consequently, the intrinsic photoelectric stopping-potential magnitude also remains unchanged: \[ V'_0=V_0 \] \(\textbf{Physical interpretation:}\) The magnetic field can prevent emitted electrons from reaching a collector through geometrical deflection, even when their kinetic energies are unchanged. A zero collected current is therefore not sufficient evidence of absent photoelectric emission or reduced photoelectron energy. Electron collection depends on trajectories as well as on initial kinetic energy. \(\textbf{Final answer:}\) The new orbit radius is \(0.75d\), so electrons cannot reach the collector; collected current falls to zero while intrinsic stopping potential remains unchanged.

Explanation: Test P is consistent because, at unchanged photon frequency and quantum efficiency, a higher incident photon arrival rate produces more emitted photoelectrons per second. With complete collection, the saturation current therefore increases. However, the maximum energy supplied by an individual photon is unchanged, so the stopping potential remains unchanged. Test Q is also consistent because photon arrival rate and quantum efficiency remain fixed. Consequently, the total emission rate and saturation current remain unchanged under the stated assumptions. The increased photon frequency raises individual photon energy and therefore increases maximum photoelectron kinetic energy. The stopping potential accordingly increases. Test R is consistent with the distinction between electron emission and electron collection. Increasing the positive collector potential can enable a larger fraction of emitted electrons to reach the collector until saturation is attained. This does not alter the kinetic energy initially acquired from photon absorption. Test S is inconsistent with ordinary single-photon photoelectric emission because each incident photon is individually below the metal's work-function threshold. Increasing the number of such photons per second does not make an individual photon sufficiently energetic to eject an electron through the assumed mechanism. The combined tests show why photon number, photon energy and collecting voltage must be controlled independently. Only the result reported in S contradicts the stated physical model.

Explanation: \(\textbf{Compare incident photon arrival rates:}\) Let the common incident optical power be \(P_{\mathrm{opt}}\). For monochromatic light: \[ R_\gamma=\frac{P_{\mathrm{opt}}}{E_\gamma} \] Thus: \[ \frac{R_{\gamma P}}{R_{\gamma Q}}=\frac{E_{\gamma Q}}{E_{\gamma P}} \] Substituting: \[ \frac{R_{\gamma P}}{R_{\gamma Q}}=\frac{6.0}{3.0} \] \[ \frac{R_{\gamma P}}{R_{\gamma Q}}=2 \] \(\textbf{Compare electron emission rates and saturation currents:}\) The quantum efficiency is the same for both beams. Therefore: \[ R_e=\eta R_\gamma \] At saturation: \[ I_s=eR_e \] Hence: \[ \frac{I_{sP}}{I_{sQ}}=\frac{R_{eP}}{R_{eQ}} \] \[ \frac{I_{sP}}{I_{sQ}}=2 \] \(\textbf{Determine individual electron kinetic energies:}\) Einstein's equation gives: \[ K_{\max}=E_\gamma-\phi \] For P: \[ K_P=3.0-2.0 \] \[ K_P=1.0\,\text{eV} \] For Q: \[ K_Q=6.0-2.0 \] \[ K_Q=4.0\,\text{eV} \] \(\textbf{Compare individual electron momenta:}\) For non-relativistic electrons: \[ p=\sqrt{2m_eK} \] Thus: \[ \frac{p_P}{p_Q}=\sqrt{\frac{K_P}{K_Q}} \] \[ \frac{p_P}{p_Q}=\sqrt{\frac14} \] \[ \frac{p_P}{p_Q}=\frac12 \] \(\textbf{Compare momentum transported per second:}\) The emitted-electron momentum rate in the common normal direction is: \[ \dot p_e=R_ep \] Therefore: \[ \frac{\dot p_{eP}}{\dot p_{eQ}}=\frac{R_{eP}}{R_{eQ}}\frac{p_P}{p_Q} \] Substituting: \[ \frac{\dot p_{eP}}{\dot p_{eQ}}=2\left(\frac12\right) \] \[ \frac{\dot p_{eP}}{\dot p_{eQ}}=1 \] The emitted-electron momentum rates are equal. \(\textbf{Compare total emitted-electron kinetic-energy powers:}\) The kinetic-energy power is: \[ P_K=R_eK \] Consequently: \[ \frac{P_{K,P}}{P_{K,Q}}=\frac{R_{eP}}{R_{eQ}}\frac{K_P}{K_Q} \] \[ \frac{P_{K,P}}{P_{K,Q}}=2\left(\frac14\right) \] \[ \frac{P_{K,P}}{P_{K,Q}}=\frac12 \] \(\textbf{Physical interpretation:}\) Beam P delivers twice as many photons per second at equal optical power, producing twice as many emitted electrons. However, each P electron has half the momentum magnitude and one-quarter the kinetic energy of a Q electron. The emitted-electron momentum rates therefore become equal, while their total kinetic-energy powers remain different. This demonstrates why photon arrival rate, electron momentum and electron kinetic energy must be compared through their distinct physical relations. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_{sP}}{I_{sQ}}=2,\qquad\frac{\dot p_{eP}}{\dot p_{eQ}}=1,\qquad\frac{P_{K,P}}{P_{K,Q}}=\frac12} \]

Explanation: \(\textbf{Interpret the retarding-current measurements:}\) At zero retarding potential, all three groups reach the collector. Therefore, the total emitted-electron current is: \[ I_s=80\,\mu\text{A} \] At a retarding magnitude of \(0.8\,\text{V}\), electrons having kinetic energy \(0.4\,\text{eV}\) cannot reach the collector. The two higher-energy groups are collected. Their combined current is: \[ I_{1.2}+I_{2.0}=50\,\mu\text{A} \] \(\textbf{Recover the lowest-energy group's contribution:}\) \[ I_{0.4}=I_s-(I_{1.2}+I_{2.0}) \] Substituting: \[ I_{0.4}=80-50 \] \[ I_{0.4}=30\,\mu\text{A} \] \(\textbf{Recover the highest-energy group's contribution:}\) At a retarding magnitude of \(1.6\,\text{V}\), only electrons having \(2.0\,\text{eV}\) can cross the barrier. Therefore: \[ I_{2.0}=20\,\mu\text{A} \] The remaining group's contribution is: \[ I_{1.2}=50-20 \] \[ I_{1.2}=30\,\mu\text{A} \] \(\textbf{Calculate the mean emitted-electron kinetic energy:}\) Because each electron carries the same charge magnitude, group currents are proportional to electron populations. Thus: \[ \overline K=\frac{I_{0.4}(0.4)+I_{1.2}(1.2)+I_{2.0}(2.0)}{I_s}\,\text{eV} \] Substituting numerical current values in microamperes: \[ \overline K=\frac{30(0.4)+30(1.2)+20(2.0)}{80}\,\text{eV} \] \[ \overline K=\frac{12+36+40}{80}\,\text{eV} \] \[ \overline K=\frac{88}{80}\,\text{eV} \] \[ \overline K=1.10\,\text{eV} \] \(\textbf{Determine total emitted-electron kinetic-energy power:}\) The total emitted-electron rate is: \[ R_e=\frac{I_s}{e} \] Therefore: \[ P_K=R_e\overline K \] Converting \(1.10\,\text{eV}\) into \(1.10e\,\text{J}\) per electron: \[ P_K=\frac{I_s}{e}(1.10e) \] \[ P_K=I_s(1.10\,\text{V}) \] Substituting: \[ P_K=(80\times10^{-6})(1.10)\,\text{W} \] \[ P_K=88\times10^{-6}\,\text{W} \] \[ P_K=88\,\mu\text{W} \] \(\textbf{Physical interpretation:}\) A retarding collector acts as an energy selector. Successive reductions in collected current reveal how many emitted electrons lie in different kinetic-energy groups. These populations permit the mean kinetic energy to be reconstructed without assuming that every emitted electron possesses the maximum kinetic energy. The total kinetic-energy power then follows from the emission rate and the mean energy. \(\textbf{Final answer:}\) \[ \boxed{\overline K=1.10\,\text{eV},\qquad P_K=88\,\mu\text{W}} \]

Explanation: A collimated beam transports a definite total number of photons per second. Initially, the beam strikes the sheet normally, so its footprint has area \(A\). When the sheet is tilted, the footprint increases because the same beam is spread across a larger surface area. The local optical power per unit sheet area therefore decreases. However, the total optical power intercepted by the sheet remains unchanged because the entire beam is still absorbed. At fixed photon frequency, the total incident photon arrival rate is consequently unchanged. The stated quantum efficiency converts the same fraction of those photons into escaping photoelectrons. Therefore, the saturation current remains unchanged after rotation. Introducing the aperture produces a different physical change. The aperture removes half the incident photons before they reach the sheet. The absorbed photon arrival rate and emitted-electron rate therefore become half their original values. Complete collection then gives half the original saturation current. Throughout the experiment, the energy of each transmitted photon remains unchanged. The metal work function is also unchanged. Consequently, the maximum initial photoelectron kinetic energy and stopping-potential magnitude remain unchanged. The experiment distinguishes local illumination intensity, total intercepted photon power and individual photon energy. A larger illuminated footprint does not necessarily mean that more photons strike the sheet each second.

Explanation: \(\textbf{Convert Einstein's equation into the plotted variables:}\) The photoelectric equation is: \[ eV_0=\frac{hc}{\lambda}-\phi \] Dividing by \(e\): \[ V_0=\frac{hc}{e\lambda}-\frac{\phi}{e} \] Multiplying by wavelength: \[ \lambda V_0=\frac{hc}{e}-\frac{\phi}{e}\lambda \] \(\textbf{Identify the slope and intercept:}\) This equation has the straight-line form: \[ y=b+s\lambda \] where: \[ y=\lambda V_0 \] The slope is: \[ s=-\frac{\phi}{e} \] The vertical intercept is: \[ b=\frac{hc}{e} \] \(\textbf{Calculate the slope from the measured points:}\) The points are: \[ (\lambda_1,y_1)=(400,440) \] \[ (\lambda_2,y_2)=(500,240) \] Therefore: \[ s=\frac{y_2-y_1}{\lambda_2-\lambda_1} \] Substituting: \[ s=\frac{240-440}{500-400} \] \[ s=\frac{-200}{100} \] \[ s=-2.0\,\text{V} \] \(\textbf{Determine the work function:}\) Since: \[ s=-\frac{\phi}{e} \] we obtain: \[ \frac{\phi}{e}=2.0\,\text{V} \] Therefore: \[ \phi=2.0\,\text{eV} \] \(\textbf{Determine the vertical intercept:}\) Using: \[ y=b+s\lambda \] at the first measured point: \[ 440=b-2.0(400) \] \[ 440=b-800 \] \[ b=1240\,\text{V nm} \] Hence: \[ \frac{hc}{e}=1240\,\text{V nm} \] \(\textbf{Determine the threshold wavelength:}\) At threshold: \[ V_0=0 \] Therefore: \[ \lambda_0V_0=0 \] Using the graph equation: \[ 0=1240-2.0\lambda_0 \] \[ 2.0\lambda_0=1240 \] \[ \lambda_0=620\,\text{nm} \] \(\textbf{Physical interpretation:}\) The stopping potential is not itself a linear function of wavelength. However, multiplying it by wavelength transforms Einstein's equation into a straight-line relationship. The slope gives the work function, while the vertical intercept contains Planck's constant and the speed of light. The zero crossing determines the physical threshold wavelength; the straight line must not be extended as a photoelectric response into the subthreshold region. \(\textbf{Final answer:}\) \[ \boxed{\phi=2.0\,\text{eV},\qquad\lambda_0=620\,\text{nm},\qquad\frac{hc}{e}=1240\,\text{V nm}} \]

Explanation: \(\textbf{Determine the uniform retarding energy gradient:}\) The full collector-emitter potential difference has magnitude: \[ V_r=3.0\,\text{V} \] The plate separation is: \[ d=6.0\,\text{mm} \] Because the field is uniform, the magnitude of the retarding potential difference over forward distance \(x\) is: \[ V_r(x)=V_r\frac{x}{d} \] The electron loses kinetic energy: \[ \Delta K(x)=eV_r\frac{x}{d} \] \(\textbf{Apply the turning-point condition:}\) At the farthest forward position, the electron's normal kinetic energy becomes zero. Therefore: \[ K_i=eV_r\frac{x_{\max}}{d} \] Rearranging: \[ x_{\max}=d\frac{K_i}{eV_r} \] \(\textbf{Calculate the initial turning distance:}\) Initially: \[ K_{i1}=1.5\,\text{eV} \] The complete retarding barrier is: \[ eV_r=3.0\,\text{eV} \] Therefore: \[ x_{\max,1}=6.0\,\text{mm}\left(\frac{1.5}{3.0}\right) \] \[ x_{\max,1}=3.0\,\text{mm} \] \(\textbf{Calculate the turning distance after frequency increase:}\) The new maximum initial energy is: \[ K_{i2}=2.4\,\text{eV} \] Hence: \[ x_{\max,2}=6.0\,\text{mm}\left(\frac{2.4}{3.0}\right) \] \[ x_{\max,2}=4.8\,\text{mm} \] \(\textbf{Test collection in both experiments:}\) The collector is located at: \[ x=d=6.0\,\text{mm} \] Both turning distances are smaller than the plate separation: \[ x_{\max,1}\lt d \] \[ x_{\max,2}\lt d \] Therefore, even the most energetic photoelectrons turn around before reaching the collector in both experiments. The collected photocurrent remains zero. \(\textbf{Physical interpretation:}\) Increasing photon frequency increases the maximum initial photoelectron kinetic energy and allows electrons to travel farther against the same electric field. However, this does not guarantee that the collector receives them. A zero photocurrent can persist under two different above-threshold frequencies because the complete retarding barrier exceeds both maximum emission energies. The turning-point position contains information that a zero-current reading alone does not provide. \(\textbf{Final answer:}\) \[ \boxed{x_{\max,1}=3.0\,\text{mm},\qquad x_{\max,2}=4.8\,\text{mm},\qquad I_1=I_2=0} \]

Explanation: \(\textbf{Determine the first source's maximum photoelectron energy:}\) The initial photon energy is: \[ E_{\gamma1}=4.0\,\text{eV} \] The work function is: \[ \phi=2.0\,\text{eV} \] Therefore: \[ K_{\max,1}=E_{\gamma1}-\phi \] \[ K_{\max,1}=4.0-2.0 \] \[ K_{\max,1}=2.0\,\text{eV} \] \(\textbf{Calculate the first electrostatic charging limit:}\) The positive stopping potential is: \[ V_1=2.0\,\text{V} \] With: \[ C=3.0\,\text{pF} \] the resulting emitter charge is: \[ Q_1=CV_1 \] \[ Q_1=(3.0\,\text{pF})(2.0\,\text{V}) \] \[ Q_1=+6.0\,\text{pC} \] \(\textbf{Determine the second source's intrinsic emission energy:}\) The replacement photons have: \[ E_{\gamma2}=3.0\,\text{eV} \] Therefore: \[ K_{\max,2}=3.0-2.0 \] \[ K_{\max,2}=1.0\,\text{eV} \] The second source could charge a previously uncharged isolated emitter towards a positive potential of \(1.0\,\text{V}\). \(\textbf{Account for the existing positive potential:}\) At the start of the second illumination, the emitter is already at: \[ V_{\mathrm{existing}}=+2.0\,\text{V} \] An electron transferred from the emitter to the grounded collector must overcome an electrostatic barrier of: \[ eV_{\mathrm{existing}}=2.0\,\text{eV} \] But: \[ K_{\max,2}=1.0\,\text{eV}\lt 2.0\,\text{eV} \] Therefore, no additional net transfer to the collector can occur under the stated model. The stored charge remains: \[ Q_2=+6.0\,\text{pC} \] \(\textbf{Apply grounding before the repeated second illumination:}\) Grounding restores: \[ Q_i=0 \] After disconnection, the second source can now transfer electrons until the emitter reaches: \[ V'_2=+1.0\,\text{V} \] Therefore: \[ Q'_2=CV'_2 \] \[ Q'_2=(3.0\,\text{pF})(1.0\,\text{V}) \] \[ Q'_2=+3.0\,\text{pC} \] \(\textbf{Physical interpretation:}\) The weaker-photon source can produce ordinary photoelectric emission because its photon energy exceeds the intrinsic work function. However, existing electrostatic charging can prevent those emitted electrons from reaching the grounded collector. Replacing illumination does not automatically discharge an isolated emitter. Grounding removes the stored electrostatic barrier and permits the second source to establish its own lower charging limit. \(\textbf{Final answer:}\) \[ \boxed{Q_2=+6.0\,\text{pC},\qquad Q'_2=+3.0\,\text{pC}} \]

Explanation: \(\textbf{Determine incident photon rate:}\) The optical power is: \[ P=1.00\times10^{-3}\,\text{W} \] Each photon carries: \[ E_\gamma=4.0\,\text{eV} \] Converting: \[ E_\gamma=4.0(1.60\times10^{-19}) \] \[ E_\gamma=6.40\times10^{-19}\,\text{J} \] Therefore: \[ R_\gamma=\frac{P}{E_\gamma} \] Substituting: \[ R_\gamma=\frac{1.00\times10^{-3}}{6.40\times10^{-19}} \] \[ R_\gamma=1.5625\times10^{15}\,\text{s}^{-1} \] \(\textbf{Determine the emitted-electron rate:}\) The saturation current is: \[ I_s=80\,\mu\text{A} \] At complete collection: \[ R_e=\frac{I_s}{e} \] Substituting: \[ R_e=\frac{80\times10^{-6}}{1.60\times10^{-19}} \] \[ R_e=5.00\times10^{14}\,\text{s}^{-1} \] \(\textbf{Calculate quantum efficiency:}\) \[ \eta=\frac{R_e}{R_\gamma} \] \[ \eta=\frac{5.00\times10^{14}}{1.5625\times10^{15}} \] \[ \eta=0.32 \] \[ \eta=32\% \] \(\textbf{Find the maximum kinetic energy per emitted electron:}\) Einstein's equation gives: \[ K_{\max}=E_\gamma-\phi \] Substituting: \[ K_{\max}=4.0-2.5 \] \[ K_{\max}=1.5\,\text{eV} \] Each emitted electron therefore carries no more than \(1.5\,\text{eV}\) of initial kinetic energy. \(\textbf{Establish the upper bound on outgoing kinetic-energy power:}\) The greatest possible total electron kinetic-energy power occurs if every emitted electron has \(K_{\max}\). Thus: \[ P_{K,e}\leq R_eK_{\max} \] Substituting: \[ P_{K,e}\leq(5.00\times10^{14})(1.5)(1.60\times10^{-19})\,\text{W} \] \[ P_{K,e}\leq1.20\times10^{-4}\,\text{W} \] \[ P_{K,e}\leq0.120\,\text{mW} \] \(\textbf{Find the minimum power not carried away as electron kinetic energy:}\) \[ P_{\mathrm{remaining}}=P-P_{K,e} \] Because \(P_{K,e}\) is bounded above: \[ P_{\mathrm{remaining}}\geq1.00-0.120 \] \[ P_{\mathrm{remaining}}\geq0.880\,\text{mW} \] \(\textbf{Interpret the bound correctly:}\) The measurements do not establish the actual distribution of emitted-electron kinetic energies. Therefore, the outgoing kinetic-energy power need not attain its calculated upper bound. The remaining power may be greater than \(0.880\,\text{mW}\), but it cannot be smaller under the stated assumptions. \(\textbf{Final answer:}\) \[ \boxed{\eta=32\%,\qquad P_{\mathrm{remaining}}\geq0.880\,\text{mW}} \]

Explanation: \(\textbf{Identify the geometric change in intercepted optical power:}\) For an isotropic point source, radiant intensity at distance \(d\) varies as: \[ J\propto\frac{1}{d^2} \] A small cathode of effective illuminated area \(A\) intercepts power proportional to: \[ P_{\mathrm{abs}}\propto\frac{A}{d^2} \] \(\textbf{Apply the changed distance and area:}\) Initially: \[ P_{\mathrm{abs},1}\propto\frac{A}{d^2} \] After modification: \[ d'=2d \] \[ A'=2A \] Therefore: \[ P_{\mathrm{abs},2}\propto\frac{2A}{(2d)^2} \] \[ P_{\mathrm{abs},2}\propto\frac{2A}{4d^2} \] Thus: \[ \frac{P_{\mathrm{abs},2}}{P_{\mathrm{abs},1}}=\frac12 \] \(\textbf{Account for the simultaneous photon-energy change:}\) The incident photon arrival rate at the cathode is: \[ R_\gamma=\frac{P_{\mathrm{abs}}}{E_\gamma} \] Therefore: \[ \frac{R_{\gamma2}}{R_{\gamma1}}=\frac{P_{\mathrm{abs},2}}{P_{\mathrm{abs},1}}\frac{E_{\gamma1}}{E_{\gamma2}} \] Substituting: \[ \frac{R_{\gamma2}}{R_{\gamma1}}=\frac12\left(\frac{3.0}{4.0}\right) \] \[ \frac{R_{\gamma2}}{R_{\gamma1}}=\frac38 \] \(\textbf{Determine the saturation-current ratio:}\) At unchanged quantum efficiency: \[ I_s=e\eta R_\gamma \] Hence: \[ \frac{I'_s}{I_s}=\frac38 \] \(\textbf{Determine the original stopping potential:}\) Einstein's equation gives: \[ K_{\max,1}=E_{\gamma1}-\phi \] \[ K_{\max,1}=3.0-2.0 \] \[ K_{\max,1}=1.0\,\text{eV} \] Therefore: \[ V_{01}=1.0\,\text{V} \] \(\textbf{Determine the modified stopping potential:}\) \[ K_{\max,2}=4.0-2.0 \] \[ K_{\max,2}=2.0\,\text{eV} \] Thus: \[ V_{02}=2.0\,\text{V} \] \(\textbf{Physical interpretation:}\) Moving the cathode farther away decreases intercepted light power. Increasing its effective area partly compensates for that loss. The higher photon energy then reduces the number of photons arriving per second for a given intercepted power. Saturation current responds to the resulting photon rate, whereas stopping potential responds to the increased energy of individual photons. \(\textbf{Final answer:}\) \[ \boxed{\frac{I'_s}{I_s}=\frac38,\qquad V_{01}=1.0\,\text{V},\qquad V_{02}=2.0\,\text{V}} \]