Dual Nature of Matter and Radiation MCQs with Answers – Part 3 (NEET Physics)

Explanation: The frequency is fixed above threshold, so each incident photon possesses the same energy before and after the change. Doubling photon arrival rate increases the number of photons reaching the emitter per second. Since quantum efficiency remains unchanged, the emitted-photoelectron rate doubles. The applied retarding potential, however, is not changed. The kinetic-energy distribution of emitted electrons is also specified to remain unchanged. Consequently, the same fraction of emitted electrons has sufficient energy and suitable direction to overcome the retarding field. The collection fraction therefore remains unchanged. The collected-electron rate equals the emission rate multiplied by the collection fraction. Doubling the first factor while keeping the second constant doubles the collected current. This conclusion applies even though the collector is not at saturation, provided the stated independent-emission and unchanged-distribution assumptions hold. It is therefore incorrect to conclude that a fixed retarding potential necessarily produces a fixed photocurrent. Retarding potential controls which fraction of emitted electrons is collected, while intensity controls how many electrons are available to be collected.

Explanation: Saturation current is obtained when essentially every emitted photoelectron reaches the collector. It therefore provides a measure of the total rate of photoelectron emission under the specified illumination conditions. At a retarding collector potential, only a fraction of the emitted electrons may reach the collector. The corresponding current measures the collected-electron rate rather than the complete emitted-electron rate. If the illumination and emission conditions are identical in the two measurements, the emitted-electron rate is common to both. Dividing the retarding-potential current by the saturation current therefore gives the collected fraction of emitted electrons. This ratio is dimensionless and lies between zero and unity under the stated ordinary conditions. However, the relation cannot be used without qualification if the source intensity, quantum efficiency or emitting surface changes between the two readings. In that situation, the saturation current would represent a different emission rate from the retarding-current measurement. The method illustrates why a saturation reference is useful when interpreting partial collection. It also separates experimental information about electron emission from information about electron transport through the collector field.

Explanation: \(\textbf{Express the work function using threshold frequency:}\) The work function is: \[ \phi=h\nu_0 \] Einstein's photoelectric equation gives: \[ K_{\max}=h\nu-\phi \] Therefore: \[ K_{\max}=h\nu-h\nu_0 \] \(\textbf{Apply the specified energy fraction:}\) The required condition is: \[ K_{\max}=0.75E_\gamma \] Since: \[ E_\gamma=h\nu \] we have: \[ h\nu-h\nu_0=0.75h\nu \] Rearranging: \[ 0.25h\nu=h\nu_0 \] Cancelling \(h\): \[ 0.25\nu=\nu_0 \] Therefore: \[ \nu=4\nu_0 \] \(\textbf{Compare photon arrival rates at constant optical power:}\) For monochromatic radiation: \[ P=R_\gamma h\nu \] Hence: \[ R_\gamma=\frac{P}{h\nu} \] Initially: \[ \nu_1=2\nu_0 \] After the change: \[ \nu_2=4\nu_0 \] Therefore: \[ \frac{R_{\gamma2}}{R_{\gamma1}}=\frac{\nu_1}{\nu_2} \] \[ \frac{R_{\gamma2}}{R_{\gamma1}}=\frac{2\nu_0}{4\nu_0} \] \[ \frac{R_{\gamma2}}{R_{\gamma1}}=\frac{1}{2} \] \(\textbf{Saturation-current comparison:}\) At constant quantum efficiency: \[ I_s=e\eta R_\gamma \] Thus: \[ \frac{I_{s2}}{I_{s1}}=\frac{1}{2} \] \(\textbf{Physical interpretation:}\) A greater fraction of individual photon energy becomes available as maximum photoelectron kinetic energy when the photon energy becomes large compared with the fixed work function. However, at constant optical power, more energetic photons arrive less frequently. Therefore, the required increase in the kinetic-energy fraction produces a smaller saturation current under the stated efficiency assumption. \(\textbf{Final answer:}\) \[ \boxed{\nu_2=4\nu_0,\qquad\frac{I_{s2}}{I_{s1}}=\frac{1}{2}} \]

Explanation: At a sufficiently positive collector potential, the electric field accelerates emitted photoelectrons towards the collector. Under saturation conditions, essentially every emitted electron reaches the collector, corresponding to stage P. As the collector potential is reduced, it eventually becomes less effective at collecting electrons emitted with different initial energies and directions. Some electrons may then fail to reach the collector, reducing the photocurrent below saturation. This corresponds to stage Q. When the collector becomes negative relative to the emitting surface, the electric field retards electrons travelling towards it. Increasing the magnitude of this retarding potential progressively prevents electrons with lower suitable kinetic-energy components from reaching the collector. At the stopping potential, even the most energetic emitted electrons cannot reach the collecting plate. The photocurrent consequently becomes zero, corresponding to stage R. The order of collection states is therefore complete collection, partial collection and no collection. Throughout the voltage sweep, suitable illumination can continue producing photoelectrons at the emitting surface. The observed current changes primarily because the collector voltage changes the collection conditions, not because it necessarily changes the photoelectron emission rate.

Explanation: \(\textbf{Maximum current corresponding to the incident photon rate:}\) The incident photon arrival rate is: \[ R_\gamma=5.0\times10^{14}\,\text{s}^{-1} \] Under the stated one-photon-one-electron limit, the greatest possible photoelectron emission rate is: \[ R_{e,\max}=R_\gamma \] Therefore, the greatest possible saturation current is: \[ I_{s,\max}=eR_\gamma \] Substituting: \[ I_{s,\max}=(1.60\times10^{-19})(5.0\times10^{14})\,\text{A} \] \[ I_{s,\max}=8.0\times10^{-5}\,\text{A} \] \[ I_{s,\max}=80\,\mu\text{A} \] \(\textbf{Separate emission probability and collection fraction:}\) Let \(\eta\) be the photoelectric quantum efficiency. Let \(f\) be the fraction of emitted electrons reaching the collector at the specified negative potential. The collected current is: \[ I=e\eta fR_\gamma \] Since collection cannot exceed complete collection: \[ 0\leq f\leq1 \] Therefore: \[ I\leq e\eta R_\gamma \] \(\textbf{Infer the efficiency bound:}\) The measured current is: \[ I=40\,\mu\text{A} \] Thus: \[ 40\,\mu\text{A}\leq\eta(80\,\mu\text{A}) \] Rearranging: \[ \eta\geq\frac{40}{80} \] \[ \eta\geq0.50 \] \(\textbf{Physical interpretation:}\) The measured current is obtained at a retarding potential rather than necessarily at saturation. Therefore, the collected-electron rate may be smaller than the emitted-electron rate. The quantum efficiency must be at least \(50\%\), but it could be larger if some emitted electrons fail to reach the collector. This provides a lower bound rather than a unique determination of quantum efficiency. \(\textbf{Final answer:}\) \[ \boxed{\eta_{\min}=50\%} \]

Explanation: The initial interference pattern results from the relative phase difference between electromagnetic waves arriving from the two slits. At P, the two contributions are initially in the constructive-interference condition, producing a bright fringe. At Q, they are initially in the destructive-interference condition, producing a dark fringe. Introducing an additional relative phase difference of \(\pi\) reverses these interference conditions at the fixed screen positions. Thus, P becomes a dark-fringe position, while Q becomes a bright-fringe position. The spatial distribution of photon detection events on the screen consequently changes. Under the stipulated ideal conditions, photoelectric emission at P is suppressed when its local intensity becomes zero, while Q can begin producing photoelectrons when its local intensity becomes nonzero. However, the optical element does not change the frequency of the monochromatic radiation. Therefore, it does not change the energy \(h\nu\) carried by individual photons. For the unchanged metallic screen, Einstein's equation predicts the same maximum photoelectron kinetic energy wherever emission occurs. Thus, the photoelectric current distribution follows the shifted interference pattern, while the energy of individual emitted photoelectrons remains governed by the original photon frequency and work function. This demonstrates how wave interference can control the spatial distribution of a photon-driven emission process.

Explanation: Statement I is correct because photon energy depends on the radiation frequency, which remains unchanged when the collector voltage is adjusted. The electrical potential of the collector acts on emitted electrons after they leave the surface rather than changing the energy carried by incoming photons. Statement II is correct because a more negative collector produces a stronger retarding effect for electrons travelling towards it. This can reduce the fraction of emitted electrons reaching the collector and hence decrease photocurrent. Such a change does not require any modification of the intrinsic work function of the unchanged emitting surface. Statement III is incorrect because zero collection current can occur even when the incident frequency is above threshold. At the stopping potential, photoelectric emission may continue while the retarding field prevents electrons from reaching the collector. Statement IV is correct when understood as the equality \(eV_0=K_{\max}\): the maximum initial photoelectron kinetic energy equals the magnitude of electrical work required to stop the most energetic emitted electrons. The statement does not mean that the collector supplied the original photoelectric kinetic energy. Rather, it means that an opposing electrical field can remove that kinetic energy during electron motion. These distinctions allow photoelectric emission, energy transfer and electron collection to be analysed separately.

Explanation: \(\textbf{Threshold restrictions from positive collection:}\) Both surfaces produce photoelectric current at sufficiently positive collector potential. For P: \[ E\gt \phi_P \] \[ E\gt 2.0\,\text{eV} \] For Q: \[ E\gt \phi_Q \] \[ E\gt 2.6\,\text{eV} \] Thus, the stronger threshold restriction is: \[ E\gt 2.6\,\text{eV} \] \(\textbf{Retarding-current observation for P:}\) The collector retarding magnitude is: \[ V_r=0.50\,\text{V} \] The required kinetic-energy barrier is: \[ \Delta U=0.50\,\text{eV} \] P produces a nonzero collected current. Therefore, its maximum photoelectron kinetic energy exceeds this barrier: \[ E-\phi_P\gt 0.50\,\text{eV} \] Substituting: \[ E-2.0\gt 0.50 \] \[ E\gt 2.50\,\text{eV} \] This is weaker than the condition \(E\gt 2.6\,\text{eV}\) already obtained from Q's emission. \(\textbf{Retarding-current observation for Q:}\) Q produces no collected current at the same retarding magnitude. Thus: \[ K_{\max,Q}\leq0.50\,\text{eV} \] Einstein's equation gives: \[ K_{\max,Q}=E-\phi_Q \] Therefore: \[ E-2.6\leq0.50 \] \[ E\leq3.10\,\text{eV} \] \(\textbf{Combine the restrictions:}\) The photoemission condition requires: \[ E\gt 2.6\,\text{eV} \] The zero-current condition for Q requires: \[ E\leq3.10\,\text{eV} \] Therefore: \[ 2.6\,\text{eV}\lt E\leq3.10\,\text{eV} \] \(\textbf{Physical interpretation:}\) A positive collecting potential establishes whether emission is possible. A retarding potential then provides additional information about emitted-electron energy. Combining both types of observation restricts the unknown photon energy more precisely than either alone. \(\textbf{Final answer:}\) \[ \boxed{2.6\,\text{eV}\lt E\leq3.10\,\text{eV}} \]

Explanation: \(\textbf{Distinguish signed collector potential from stopping-potential magnitude:}\) Einstein's equation gives: \[ eV_0=h\nu-\phi \] Here \(V_0\) is a positive magnitude. The actual collector potential at stopping is negative relative to the emitter: \[ V_C=-V_0 \] Therefore: \[ V_C=-\frac{h}{e}\nu+\frac{\phi}{e} \] \(\textbf{Interpret the graph slope:}\) The slope of \(V_C\) against \(\nu\) is: \[ m=-\frac{h}{e} \] It must be negative because a greater incident frequency requires a more negative collector potential to stop the more energetic photoelectrons. \(\textbf{Calculate the experimental slope:}\) The measured points are: \[ (\nu_1,V_{C1})=(6.0\times10^{14}\,\text{Hz},-0.50\,\text{V}) \] \[ (\nu_2,V_{C2})=(1.0\times10^{15}\,\text{Hz},-2.16\,\text{V}) \] Thus: \[ m=\frac{V_{C2}-V_{C1}}{\nu_2-\nu_1} \] Substituting: \[ m=\frac{-2.16-(-0.50)}{1.0\times10^{15}-6.0\times10^{14}} \] \[ m=\frac{-1.66}{4.0\times10^{14}}\,\text{V s} \] \[ m=-4.15\times10^{-15}\,\text{V s} \] \(\textbf{Determine Planck's constant:}\) Since: \[ m=-\frac{h}{e} \] we have: \[ h=-em \] Substituting: \[ h=-(1.60\times10^{-19})(-4.15\times10^{-15})\,\text{C V s} \] \[ h=6.64\times10^{-34}\,\text{J s} \] \(\textbf{Physical interpretation:}\) Plotting the signed stopping collector voltage reverses the graph's vertical direction compared with plotting the positive stopping-potential magnitude. The slope changes sign, but its magnitude still determines Planck's constant. Confusing a signed voltage with its magnitude would therefore produce an incorrect interpretation of the graph. \(\textbf{Final answer:}\) \[ \boxed{h=6.64\times10^{-34}\,\text{J s},\qquad m\lt 0} \]

Explanation: \(\textbf{Initial current during illumination:}\) The incident photon arrival rate during the initial ON interval is: \[ R_{\gamma1}=R \] At saturation: \[ I_{s1}=e\eta R \] \(\textbf{Initial charge collected per cycle:}\) The initial ON duration is: \[ t_1=2\,\text{ms} \] The collected charge per cycle is: \[ Q_1=I_{s1}t_1 \] \[ Q_1=e\eta R(2\,\text{ms}) \] \(\textbf{Changed illumination rate:}\) During the new ON interval: \[ R_{\gamma2}=\frac{2R}{5} \] Therefore: \[ I_{s2}=e\eta\frac{2R}{5} \] Taking the current ratio: \[ \frac{I_{s2}}{I_{s1}}=\frac{2}{5} \] \(\textbf{Changed charge collected per cycle:}\) The new ON duration is: \[ t_2=5\,\text{ms} \] Hence: \[ Q_2=I_{s2}t_2 \] \[ Q_2=e\eta\frac{2R}{5}(5\,\text{ms}) \] \[ Q_2=e\eta R(2\,\text{ms}) \] Therefore: \[ Q_2=Q_1 \] \(\textbf{Stopping-potential comparison:}\) The maximum photoelectron kinetic energy is: \[ K_{\max}=h\nu-\phi \] The frequency and emitting surface are unchanged. Consequently: \[ V_{02}=V_{01} \] \(\textbf{Physical interpretation:}\) The modified pulse has a smaller instantaneous photon arrival rate but a longer illumination duration. These two changes compensate exactly in the total photon count per cycle. Consequently, the total collected charge per cycle remains unchanged even though the instantaneous current during illumination decreases. The stopping potential remains unchanged because the photon energy and work function do not change. This distinguishes instantaneous current, integrated charge and photoelectron energy in a pulsed photoelectric experiment. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_{s2}}{I_{s1}}=\frac{2}{5},\qquad\frac{Q_2}{Q_1}=1,\qquad\frac{V_{02}}{V_{01}}=1} \]

Explanation: \(\textbf{Infer the optical-power ratio from photon momentum transfer:}\) For a perfectly absorbing target under normal incidence, the momentum transferred per second is: \[ \dot p=\frac{P_{\text{opt}}}{c} \] The measured momentum-transfer ratio is: \[ \frac{\dot p_P}{\dot p_Q}=2 \] Therefore: \[ \frac{P_P}{P_Q}=2 \] \(\textbf{Compare individual photon energies:}\) For radiation in vacuum: \[ E_\gamma=\frac{hc}{\lambda} \] Hence: \[ \frac{E_P}{E_Q}=\frac{\lambda_Q}{\lambda_P} \] Substituting: \[ \frac{E_P}{E_Q}=\frac{600}{400} \] \[ \frac{E_P}{E_Q}=\frac{3}{2} \] \(\textbf{Determine the photon arrival-rate ratio:}\) Optical power satisfies: \[ P_{\text{opt}}=R_\gamma E_\gamma \] Therefore: \[ \frac{R_P}{R_Q}=\frac{P_P}{P_Q}\frac{E_Q}{E_P} \] Substituting: \[ \frac{R_P}{R_Q}=2\times\frac{2}{3} \] \[ \frac{R_P}{R_Q}=\frac{4}{3} \] \(\textbf{Apply the equal-saturation-current condition:}\) At saturation: \[ I_s=e\eta R_\gamma \] The two measured saturation currents are equal: \[ e\eta_PR_P=e\eta_QR_Q \] Consequently: \[ \frac{\eta_P}{\eta_Q}=\frac{R_Q}{R_P} \] \[ \frac{\eta_P}{\eta_Q}=\frac{3}{4} \] \(\textbf{Physical interpretation:}\) Beam P has twice the optical power but its photons are individually more energetic. Its photon arrival rate is therefore only \(4/3\) times that of Q. To produce the same saturation current, P must have a correspondingly smaller photoelectric quantum efficiency. The experiment combines momentum-transfer measurements with photoelectron counting to infer an emission property that neither measurement determines independently. \(\textbf{Final answer:}\) \[ \boxed{\frac{\eta_P}{\eta_Q}=\frac{3}{4}} \]

Explanation: \(\textbf{Identify the threshold restriction:}\) The metal work function is: \[ \phi=2.0\,\text{eV} \] Ordinary photoelectric emission requires: \[ E_\gamma\geq\phi \] Source P has photon energy below the work function, so it cannot produce the measured nonzero photoelectric current. \(\textbf{Find the maximum current permitted by a photon energy:}\) If the incident optical power is \(P_{\text{opt}}\), the photon arrival rate is: \[ R_\gamma=\frac{P_{\text{opt}}}{E_\gamma} \] With at most one electron emitted per photon: \[ I_{s,\max}=eR_\gamma \] Therefore: \[ I_{s,\max}=\frac{eP_{\text{opt}}}{E_\gamma} \] When photon energy is expressed in electronvolts, the numerical energy value can be used as an equivalent voltage: \[ I_{s,\max}=\frac{P_{\text{opt}}}{E_\gamma/e} \] \(\textbf{Test source Q:}\) For Q: \[ E_Q=2.5\,\text{eV} \] Thus: \[ I_{s,\max,Q}=\frac{0.80\times10^{-3}\,\text{W}}{2.5\,\text{V}} \] \[ I_{s,\max,Q}=3.20\times10^{-4}\,\text{A} \] \[ I_{s,\max,Q}=320\,\mu\text{A} \] The measured current is possible if the photoelectric quantum efficiency is unity. \(\textbf{Test source R:}\) For R: \[ I_{s,\max,R}=\frac{0.80\times10^{-3}}{3.0}\,\text{A} \] \[ I_{s,\max,R}\approx267\,\mu\text{A} \] This is smaller than the measured current. \(\textbf{Test source S:}\) For S: \[ I_{s,\max,S}=\frac{0.80\times10^{-3}}{4.0}\,\text{A} \] \[ I_{s,\max,S}=200\,\mu\text{A} \] This is also smaller than the measured current. \(\textbf{Consistency conclusion:}\) Source P fails the emission-threshold condition. Sources R and S have too few incident photons per second to produce the recorded current under the specified one-electron-per-photon limit. Source Q alone satisfies both requirements. \(\textbf{Final answer:}\) Only source Q is consistent with the measurement.

Explanation: \(\textbf{Compare photon numbers in equal-energy pulses:}\) Let the total radiant energy in each pulse be \(U\). The number of photons is: \[ N=\frac{U}{E_\gamma} \] For P: \[ N_P=\frac{U}{3.0\,\text{eV}} \] For Q: \[ N_Q=\frac{U}{4.5\,\text{eV}} \] Therefore: \[ \frac{N_P}{N_Q}=\frac{4.5}{3.0} \] \[ \frac{N_P}{N_Q}=\frac{3}{2} \] \(\textbf{Compare collected photoelectric charges:}\) Let the common photoelectric quantum efficiency be \(\eta\). Under complete collection: \[ Q_{\text{collected}}=e\eta N \] Hence: \[ \frac{Q_P}{Q_Q}=\frac{N_P}{N_Q} \] \[ \frac{Q_P}{Q_Q}=\frac{3}{2} \] \(\textbf{Compare momentum delivered during absorption:}\) Each photon carries momentum magnitude: \[ p_\gamma=\frac{E_\gamma}{c} \] For a unidirectional pulse whose photons are all absorbed: \[ p_{\text{total}}=Np_\gamma \] Substituting the photon number: \[ p_{\text{total}}=\frac{U}{E_\gamma}\frac{E_\gamma}{c} \] \[ p_{\text{total}}=\frac{U}{c} \] Since the pulses carry equal total energies: \[ p_{\text{total},P}=p_{\text{total},Q} \] \(\textbf{Physical interpretation:}\) Equal total radiant energy does not imply equal photon number. The lower-energy pulse contains more photons and therefore produces more emitted electrons when quantum efficiency is unchanged. However, every photon contributes momentum proportional to its energy, so the total incident momentum in each unidirectional pulse depends only on the total energy. The two measurements therefore reveal different aspects of the same radiation. \(\textbf{Final answer:}\) \[ \boxed{\frac{Q_P}{Q_Q}=\frac{3}{2},\qquad\frac{p_{\text{total},P}}{p_{\text{total},Q}}=1} \]

Explanation: Statement I is correct because the two beams have equal optical powers and propagate in opposite directions. For photons in vacuum, momentum magnitude per unit time equals optical power divided by the speed of light. Their equal and opposite incident momentum rates therefore cancel vectorially. Statement II is correct because both beams have the same wavelength and optical power. Each beam consequently supplies the same number of photons per second. Illuminating the sheet with both beams doubles the total incident photon arrival rate relative to either beam alone. Statement III is correct because the two faces have equal photoelectric quantum efficiencies and their emitted electrons are collected separately. Each beam produces the same emitted-electron rate under the stated conditions, so the total saturation-current magnitude is twice the single-beam contribution. Statement IV is incorrect because zero resultant momentum does not mean that the incident photons carry zero energy or that no photons strike the surface. The radiant energies delivered from the two directions add, whereas the incident momenta cancel. Suitable photons can therefore produce photoelectric emission from both faces even when the total incident photon momentum vector is zero. This distinction also illustrates why energy, photon count and momentum must be combined according to their different physical properties. Energy and count are additive scalar quantities, while momentum is a vector.

Explanation: \(\textbf{Energy needed for photoelectron collection:}\) An incident photon must first supply the metal's work-function energy: \[ \phi=2.2\,\text{eV} \] The maximum photoelectron kinetic energy is: \[ K_{\max}=E_\gamma-\phi \] The collector provides a retarding barrier: \[ \Delta U=eV_r \] For: \[ V_r=0.60\,\text{V} \] the barrier is: \[ \Delta U=0.60\,\text{eV} \] \(\textbf{Determine the critical incident photon energy:}\) At the critical boundary: \[ K_{\max}=\Delta U \] Therefore: \[ E_{\gamma,\text{crit}}-\phi=eV_r \] Rearranging: \[ E_{\gamma,\text{crit}}=\phi+eV_r \] Substituting: \[ E_{\gamma,\text{crit}}=2.2+0.60 \] \[ E_{\gamma,\text{crit}}=2.80\,\text{eV} \] \(\textbf{Convert energy into photon momentum:}\) For photons in vacuum: \[ E_\gamma=p_\gamma c \] Hence: \[ p_{\text{crit}}=\frac{E_{\gamma,\text{crit}}}{c} \] Convert energy into joules: \[ E_{\gamma,\text{crit}}=2.80(1.60\times10^{-19})\,\text{J} \] \[ E_{\gamma,\text{crit}}=4.48\times10^{-19}\,\text{J} \] Substituting: \[ p_{\text{crit}}=\frac{4.48\times10^{-19}}{3.0\times10^8} \] \[ p_{\text{crit}}=1.493\times10^{-27}\,\text{kg m s}^{-1} \] \(\textbf{Distinguish emission from collection:}\) Photon energy greater than the work function is sufficient for photoelectric emission. However, the emitted electrons must additionally overcome the retarding barrier to reach the collector. At the critical momentum, even the most energetic electrons just fail to cross the barrier in the ideal limiting condition. A nonzero collected current is possible when the photon momentum is strictly greater than this critical value. \(\textbf{Final answer:}\) \[ \boxed{p_{\text{crit}}\approx1.49\times10^{-27}\,\text{kg m s}^{-1},\qquad p_\gamma\gt p_{\text{crit}}\ \text{for nonzero collection}} \]

Explanation: The assertion is correct because the measured photocurrent depends on the number of emitted electrons that successfully reach the collector. Photoelectric emission is caused by suitable incident photons transferring energy to electrons at the metal surface. The photon frequency and arrival rate are unchanged, so the ordinary emission conditions remain unchanged under the stated assumptions. However, the emitted particles are electrons and therefore carry negative electric charge. When they move through a magnetic field with a velocity component perpendicular to the field, they experience a magnetic force. This force is perpendicular to the instantaneous velocity and can bend the electron trajectory. Because the collector has finite dimensions, a deflected electron may miss it even if the electron initially travelled towards it. The measured current can consequently decrease without a corresponding decrease in the number of electrons emitted per second. The magnetic force itself does no work, so it does not change the electron's kinetic energy. The reason correctly identifies the mechanism responsible for the altered collection efficiency. This experiment illustrates why a reduced measured photocurrent does not always imply reduced photon absorption or photoelectric emission. Electron emission and electron transport must be distinguished when external fields influence the collection process.

Explanation: \(\textbf{Distinguish the two efficiencies:}\) Photoelectric quantum efficiency is: \[ \eta=\frac{R_e}{R_\gamma} \] where \(R_e\) is the emitted-electron rate and \(R_\gamma\) is the incident photon arrival rate. Collection efficiency is: \[ f=\frac{R_c}{R_e} \] where \(R_c\) is the collected-electron rate. \(\textbf{Evaluate condition P:}\) \[ \eta_P=\frac{0.4R}{R} \] \[ \eta_P=0.40 \] The collection fraction is: \[ f_P=\frac{0.3R}{0.4R} \] \[ f_P=0.75 \] \(\textbf{Evaluate condition Q:}\) \[ \eta_Q=\frac{0.6R}{2R} \] \[ \eta_Q=0.30 \] The collection fraction is: \[ f_Q=\frac{0.3R}{0.6R} \] \[ f_Q=0.50 \] \(\textbf{Evaluate condition S:}\) \[ \eta_S=\frac{1.2R}{3R} \] \[ \eta_S=0.40 \] The collection fraction is: \[ f_S=\frac{0.6R}{1.2R} \] \[ f_S=0.50 \] \(\textbf{Compare the results:}\) Conditions P and S have equal quantum efficiency: \[ \eta_P=\eta_S=0.40 \] But: \[ f_P=0.75\gt f_S=0.50 \] Condition Q and S have equal collection efficiency but different quantum efficiencies. \(\textbf{Physical interpretation:}\) The emitted-electron rate measures the success of the photoelectric emission process, while the collected-electron rate additionally depends on electron transport to the collector. Equal collected currents therefore need not imply equal quantum efficiencies. The table allows the two efficiencies to be separated because all three rates are supplied independently. \(\textbf{Final answer:}\) \[ \boxed{\eta_P=\eta_S=0.40,\qquad f_P=0.75\gt f_S=0.50} \]

Explanation: \(\textbf{Interpret the photon-momentum graph:}\) Photon momentum in vacuum is: \[ p_\gamma=\frac{h\nu}{c} \] Therefore: \[ p_\gamma=\left(\frac{h}{c}\right)\nu \] The graph slope is: \[ s_P=\frac{h}{c} \] \(\textbf{Interpret the photoelectric stopping-potential graph:}\) Einstein's equation gives: \[ eV_0=h\nu-\phi \] Rearranging: \[ V_0=\frac{h}{e}\nu-\frac{\phi}{e} \] Hence, the stopping-potential graph slope is: \[ s_Q=\frac{h}{e} \] \(\textbf{Eliminate Planck's constant:}\) From the first experiment: \[ h=cs_P \] From the second experiment: \[ h=es_Q \] Equating: \[ cs_P=es_Q \] Therefore: \[ e=\frac{cs_P}{s_Q} \] \(\textbf{Substitute the measured values:}\) \[ e=\frac{(3.0\times10^8)(2.21\times10^{-42})}{4.14\times10^{-15}} \] The numerator is: \[ cs_P=6.63\times10^{-34}\,\text{J s} \] Therefore: \[ e=\frac{6.63\times10^{-34}}{4.14\times10^{-15}}\,\text{C} \] \[ e=1.601\times10^{-19}\,\text{C} \] \(\textbf{Physical interpretation:}\) The two graphs involve the same universal Planck constant but different physical quantities on their vertical axes. The photon-momentum graph has slope \(h/c\), while the stopping-potential graph has slope \(h/e\). Dividing the slopes appropriately removes \(h\), allowing the electron charge magnitude to be inferred from two independent radiation measurements. \(\textbf{Final answer:}\) \[ \boxed{e\approx1.60\times10^{-19}\,\text{C}} \]

Explanation: \(\textbf{Compare saturation currents:}\) Let the equal incident photon rates be \(R\). At saturation: \[ I_s=e\eta R \] For the first beam: \[ I_{s1}=e\eta R \] For the second beam: \[ I_{s2}=e(2\eta)R \] Therefore: \[ \frac{I_{s2}}{I_{s1}}=2 \] \(\textbf{Compare maximum photoelectron kinetic energies:}\) Einstein's equation gives: \[ K_{\max}=E_\gamma-\phi \] For the first beam: \[ K_1=3.0-2.0 \] \[ K_1=1.0\,\text{eV} \] For the second beam: \[ K_2=4.0-2.0 \] \[ K_2=2.0\,\text{eV} \] Since: \[ eV_0=K_{\max} \] we obtain: \[ \frac{V_{02}}{V_{01}}=2 \] \(\textbf{Compare incident photon momentum rates:}\) The momentum carried by a photon is: \[ p_\gamma=\frac{E_\gamma}{c} \] With the same photon arrival rate \(R\), the incident photon momentum magnitude per second is: \[ \dot p=R\frac{E_\gamma}{c} \] Therefore: \[ \frac{\dot p_2}{\dot p_1}=\frac{E_2}{E_1} \] \[ \frac{\dot p_2}{\dot p_1}=\frac{4.0}{3.0} \] \[ \frac{\dot p_2}{\dot p_1}=\frac{4}{3} \] \(\textbf{Physical interpretation:}\) The current doubles because the photoelectric quantum efficiency doubles at unchanged photon arrival rate. The stopping potential doubles because the excess photon energy above the work function doubles. The photon momentum rate changes by a different factor because it depends directly on photon energy, not on the work function or emission efficiency. These three comparisons arise from three distinct physical relationships. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_{s2}}{I_{s1}}=2,\qquad\frac{V_{02}}{V_{01}}=2,\qquad\frac{\dot p_2}{\dot p_1}=\frac{4}{3}} \]

Explanation: \(\textbf{Maximum photoelectron kinetic energy:}\) The incident photon energy is: \[ E_\gamma=4.0\,\text{eV} \] The work function is: \[ \phi=2.0\,\text{eV} \] Einstein's equation gives: \[ K_{\max}=E_\gamma-\phi \] \[ K_{\max}=4.0-2.0 \] \[ K_{\max}=2.0\,\text{eV} \] \(\textbf{Determine the emitted-electron rate:}\) The saturation current is: \[ I_s=80\times10^{-6}\,\text{A} \] At saturation: \[ I_s=eR_e \] Therefore: \[ R_e=\frac{I_s}{e} \] Substituting: \[ R_e=\frac{80\times10^{-6}}{1.60\times10^{-19}}\,\text{s}^{-1} \] \[ R_e=5.0\times10^{14}\,\text{s}^{-1} \] \(\textbf{Use the measured kinetic-energy transport rate:}\) The total initial kinetic-energy power is: \[ P_K=120\,\mu\text{J s}^{-1} \] \[ P_K=120\times10^{-6}\,\text{W} \] The mean kinetic energy per emitted electron is: \[ \overline K=\frac{P_K}{R_e} \] Substituting: \[ \overline K=\frac{120\times10^{-6}}{5.0\times10^{14}}\,\text{J} \] \[ \overline K=2.40\times10^{-19}\,\text{J} \] Converting into electronvolts: \[ \overline K=\frac{2.40\times10^{-19}}{1.60\times10^{-19}}\,\text{eV} \] \[ \overline K=1.50\,\text{eV} \] \(\textbf{Compare mean and maximum kinetic energies:}\) \[ \frac{\overline K}{K_{\max}}=\frac{1.50}{2.00} \] \[ \frac{\overline K}{K_{\max}}=\frac{3}{4} \] \(\textbf{Physical interpretation:}\) The stopping-potential concept determines the maximum energy of an emitted photoelectron, whereas the measured kinetic-energy power and saturation current together determine the mean energy. These quantities need not be equal because photoelectrons can leave the metal with different kinetic energies. The result is consistent with the mean energy being smaller than the maximum energy. \(\textbf{Final answer:}\) \[ \boxed{\overline K=1.50\,\text{eV},\qquad\frac{\overline K}{K_{\max}}=\frac{3}{4}} \]