Dual Nature of Matter and Radiation MCQs with Answers – Part 2 (NEET Physics)

Explanation: \(\textbf{Threshold-energy condition:}\) For the detector-gap metal, the minimum photon energy required for ordinary photoelectric emission is: \[ \phi=h\nu_0 \] A photon can liberate an electron only when: \[ h\nu\geq h\nu_0 \] \(\textbf{Evaluate source P:}\) Its photon frequency is: \[ \nu_P=0.8\nu_0 \] Therefore: \[ E_P=h(0.8\nu_0) \] \[ E_P=0.8\phi \] Since: \[ E_P\lt \phi \] individual photons from P cannot liberate electrons through the assumed ordinary photoelectric mechanism. The larger radiant power means that more radiant energy is delivered per second, but it does not increase the energy of an individual photon. \(\textbf{Evaluate source Q:}\) Its photon frequency is: \[ \nu_Q=1.5\nu_0 \] Hence: \[ E_Q=1.5h\nu_0 \] \[ E_Q=1.5\phi \] Since: \[ E_Q\gt \phi \] photons from Q can produce photoelectric emission. \(\textbf{Interpret the spark observation:}\) Under the stated assumption, electrons liberated by Q can facilitate the ultraviolet-induced enhancement of spark discharge. Source P cannot contribute to this enhancement through ordinary photoelectric emission, irrespective of its greater power. The comparison does not predict an exact spark intensity because that would require additional discharge information. \(\textbf{Final answer:}\) Source Q can produce the assumed photoelectric spark enhancement, whereas source P cannot.

Explanation: In Lenard's evacuated-tube arrangement, suitable ultraviolet radiation falling on the emitter causes electrons to escape from its surface. A collector maintained at a positive potential relative to the emitter attracts these negatively charged photoelectrons. Their movement through the evacuated space and the accompanying charge transport through the external circuit produce the measured photocurrent. Thus, photocurrent flows during the first operation. When ultraviolet radiation is completely blocked, the source of photoelectric electron emission is removed. Under the stated steady conditions, photoelectric emission stops and the associated current becomes zero. The collector voltage alone cannot continuously produce a photoelectric current when the required illumination is absent. Restoring suitable ultraviolet illumination once again enables electron emission. The positive collector can then collect the newly emitted electrons, restoring the photocurrent. This sequence demonstrates the dependence of photoelectric current on incident radiation. It also distinguishes a sustained illumination-dependent current from a temporary electrical transient that may arise when the experimental conditions are changed.

Explanation: Ultraviolet illumination can cause electrons to escape from the zinc surface when the incident photons possess sufficient energy. Electrons carry negative electric charge, so their removal increases the net positive charge of the remaining metal. If the plate is initially negatively charged, it contains excess electrons. Losing some of these electrons reduces the magnitude of its negative charge. The plate therefore becomes less negative and may eventually become neutral or positive if sufficient electrons are removed. If the plate is initially neutral, the number of positive and negative charges is balanced. Emission of electrons disturbs this balance and leaves the plate with a positive net charge. If the plate is already positively charged, further electron removal increases its positive charge, provided emission continues. The three cases are therefore governed by the same charge-transfer mechanism. The direction of charge change does not reverse merely because the initial charge of the plate is different. These observations provide evidence that negatively charged particles leave the illuminated metal. They are also consistent with charge conservation because the emitted electrons carry away the negative charge lost by the plate.

Explanation: \(\textbf{Initial and intermediate plate charges:}\) Initially: \[ Q_i=-2.4\,\text{nC} \] After the first illumination stage: \[ Q_1=+0.8\,\text{nC} \] When electrons escape, the plate loses negative charge and its net charge increases. \(\textbf{Charge change during the first stage:}\) \[ \Delta Q_1=Q_1-Q_i \] Substituting: \[ \Delta Q_1=0.8-(-2.4) \] \[ \Delta Q_1=3.2\,\text{nC} \] Converting to coulombs: \[ \Delta Q_1=3.2\times10^{-9}\,\text{C} \] \(\textbf{Number of electrons emitted:}\) Each escaped electron increases the plate charge by \(e\). Therefore: \[ N_1=\frac{\Delta Q_1}{e} \] Substituting: \[ N_1=\frac{3.2\times10^{-9}}{1.60\times10^{-19}} \] \[ N_1=2.0\times10^{10} \] \(\textbf{Second emission stage:}\) The additional number of emitted electrons is: \[ N_2=\frac{N_1}{4} \] \[ N_2=\frac{2.0\times10^{10}}{4} \] \[ N_2=5.0\times10^9 \] The corresponding increase in plate charge is: \[ \Delta Q_2=N_2e \] \[ \Delta Q_2=(5.0\times10^9)(1.60\times10^{-19})\,\text{C} \] \[ \Delta Q_2=0.8\times10^{-9}\,\text{C} \] \[ \Delta Q_2=0.8\,\text{nC} \] \(\textbf{Final plate charge:}\) \[ Q_f=Q_1+\Delta Q_2 \] \[ Q_f=0.8+0.8 \] \[ Q_f=+1.6\,\text{nC} \] \(\textbf{Physical check:}\) The charge becomes more positive during both stages because electrons are removed from the plate. The sign of the initial charge does not change the amount of positive charge left behind by each escaped electron. The final positive charge is therefore consistent with the charge-conservation mechanism underlying Hallwachs' observations. \(\textbf{Final answer:}\) \[ \boxed{N_1=2.0\times10^{10},\qquad Q_f=+1.6\,\text{nC}} \]

Explanation: Statement I is correct because the illuminated emitter C releases photoelectrons into the evacuated tube. These electrons travel towards the positively maintained collector A under the influence of the electric field. Statement II is also correct because conventional electric current is defined in the direction in which positive charge would move. Electron motion represents negative-charge transport, so the corresponding conventional-current direction is opposite to the direction of electron movement. Thus, in the evacuated space, conventional current is directed from A towards C. Statement III is incorrect because incident photons carry energy and momentum but possess no electric charge. Photons do not themselves transport negative charge between the electrodes. The transported negative charge belongs to the emitted electrons. Statement IV is correct because an electron experiences an electric force opposite to the direction of the electric field. With A maintained at a higher potential than C, the electric force accelerates electrons towards the collector. The experiment therefore involves distinct physical processes: absorption of radiation, emission of electrons and transport of electric charge. Keeping these processes separate prevents confusion between photon propagation, electron motion and conventional current direction.

Explanation: \(\textbf{Energy of the longer-wavelength radiation:}\) For electromagnetic radiation in vacuum: \[ E_\gamma=\frac{hc}{\lambda} \] At wavelength \(600\,\text{nm}\): \[ E_{600}=\frac{1240}{600}\,\text{eV} \] \[ E_{600}=2.0667\,\text{eV} \] \[ E_{600}\approx2.07\,\text{eV} \] \(\textbf{Energy of the shorter-wavelength radiation:}\) At wavelength \(300\,\text{nm}\): \[ E_{300}=\frac{1240}{300}\,\text{eV} \] \[ E_{300}=4.1333\,\text{eV} \] \[ E_{300}\approx4.13\,\text{eV} \] \(\textbf{Infer the work function of P:}\) Metal P emits photoelectrons under \(600\,\text{nm}\) illumination. Therefore: \[ \phi_P\leq E_{600} \] Using the calculated energy: \[ \phi_P\leq2.07\,\text{eV} \] The \(300\,\text{nm}\) photons have greater energy, so their ability to produce emission is consistent with this restriction. \(\textbf{Infer the work function of Q:}\) Metal Q does not emit under \(600\,\text{nm}\) illumination. Therefore: \[ \phi_Q\gt E_{600} \] However, it emits under \(300\,\text{nm}\) illumination. Consequently: \[ \phi_Q\leq E_{300} \] Combining: \[ 2.07\,\text{eV}\lt \phi_Q\leq4.13\,\text{eV} \] \(\textbf{Physical interpretation:}\) Metal P has a sufficiently small work function to respond to the longer-wavelength radiation. Metal Q requires higher photon energy and therefore responds only to the shorter wavelength in this experiment. This is consistent with the observation that certain alkali metals can respond to visible radiation, while metals such as zinc generally require higher-frequency ultraviolet radiation. \(\textbf{Final answer:}\) \[ \boxed{\phi_P\leq2.07\,\text{eV},\qquad2.07\,\text{eV}\lt \phi_Q\leq4.13\,\text{eV}} \]

Explanation: The photoelectric tube is designed so that radiation capable of producing electron emission can reach the photosensitive emitter. In the original arrangement, the quartz window transmits the relevant ultraviolet radiation. These photons can then transfer energy to electrons in the emitting surface. If their energies exceed the escape requirement, photoelectrons are emitted and may be collected by the positively maintained collector. Ordinary glass can strongly absorb ultraviolet radiation even when it transmits a substantial amount of visible light. The replacement window therefore prevents the required ultraviolet photons from reaching the emitter under the stated conditions. Since the photoelectric emission mechanism depends on photons actually incident on the metal, maintaining the same ultraviolet source outside the tube is insufficient. The absorbed radiation transfers energy to the glass rather than to the emitting surface. Consequently, the photoelectric emission rate and associated steady photocurrent become zero. Changing the window does not increase photon energy or reduce the work function of the emitter. This experiment demonstrates why the optical transmission properties of the apparatus must be considered when interpreting photoelectric measurements.

Explanation: Experiment P involves an isolated metallic plate, whereas experiment Q involves an emitter and collector connected through an external circuit. In Experiment P, suitable ultraviolet photons can cause electrons to escape from the zinc surface. Because the plate is electrically isolated, electrons removed from it are not automatically replaced through an external electrical connection. The remaining plate therefore acquires positive net charge. An electroscope can indicate this change in charge, but there is no sustained external photocurrent through a circuit connected to the isolated plate. In Experiment Q, suitable illumination also liberates electrons from the emitting surface. However, the positive collector attracts the emitted electrons across the evacuated region. The external circuit provides a path for sustained charge transport and can register a steady photocurrent under suitable operating conditions. Thus, both experiments demonstrate radiation-induced electron emission, but the electrical quantities being observed differ. Hallwachs' arrangement emphasises change in the net charge of an isolated metal, while Lenard's arrangement permits measurement of electron collection as electric current. The distinction does not imply different charge carriers or different fundamental emission mechanisms. It reflects the electrical boundary conditions imposed by the two experimental arrangements.

Explanation: Photoelectric emission requires suitable incident radiation to transfer sufficient energy to electrons at the emitter surface. When the ultraviolet source is switched on, above-threshold photons reach the metal and can liberate electrons. The positively maintained collector attracts these electrons and enables an external photocurrent. With the source switched off, the supply of photons responsible for emission ceases. Under the stated negligible-delay conditions, the emission rate and associated steady photocurrent therefore fall to zero during dark intervals. A fixed positive collector potential does not independently create photoelectrons from the metal. Its role is to collect electrons that have already been emitted. The photon description also explains why the current does not require a long energy-accumulation interval after each illumination switch-on. An individual sufficiently energetic photon can cause emission during an appropriate interaction. The collected current consequently follows changes in illumination rather than depending on the total optical energy accumulated over many previous cycles. The assumption of negligible charge storage excludes persistent currents caused merely by delayed electrical discharge of the measuring apparatus. The observed switching response is therefore a direct test of the illumination dependence of photoelectric emission.

Explanation: The absence of photocurrent may arise either because electrons are not emitted or because emitted electrons fail to reach the collector. In this experiment, the collector is already maintained at a sufficiently positive potential for complete collection, which reduces the likelihood of a collection-related explanation. The next important question is whether the incident photons possess enough energy to overcome the work function of emitter Q. Photon energy is proportional to incident frequency. Increasing frequency therefore provides a direct way to test whether the original radiation was below the threshold frequency of Q. If emission begins after the frequency crosses a definite threshold, the observation supports an energy-threshold explanation. Merely increasing intensity at unchanged subthreshold frequency cannot make the individual photons sufficiently energetic in the ordinary single-photon process. Similarly, extending illumination duration does not remove the threshold requirement. Increasing collector voltage does not change the photon energy absorbed by electrons at the emitting surface. Changing the illuminated area also fails to establish whether photon energy exceeds the work function. Therefore, a controlled increase in frequency is the most direct test of the proposed explanation. The collector conditions should remain suitable so that any newly emitted photoelectrons can be detected.

Explanation: The assertion is correct because electron emission changes the net charge of an electrically isolated metal plate. Initially, a negatively charged zinc plate contains excess electrons. Suitable ultraviolet illumination causes electrons to escape, making the plate progressively less negative. Continued electron emission can leave the plate positively charged. Once the plate acquires positive charge, it produces an electric field that attracts negatively charged electrons. An electron attempting to escape from the plate must then move against this electrostatic attraction. The resulting electrostatic energy requirement can oppose further electron escape. This effect is distinct from a change in the frequency of the incident ultraviolet radiation. The photon energy remains fixed because the source frequency is unchanged. Nevertheless, the increasing positive charge can alter the ability of emitted electrons to escape to the surroundings. The reason correctly identifies the electric attraction responsible for this effect. Thus, an isolated illuminated plate need not continue losing electrons at the same rate indefinitely. This also illustrates why the electrical conditions surrounding an emitting surface matter when interpreting charge-accumulation experiments.

Explanation: \(\textbf{Compare emitted-electron rates:}\) Let the equal photon arrival rates be \(R\), and let the common quantum efficiency be \(\eta\). At saturation: \[ I_s=e\eta R \] Since both \(R\) and \(\eta\) are identical: \[ I_{sQ}=I_{sP} \] \(\textbf{Photoelectron kinetic energy under P:}\) Let the metal work function be \(\phi\). Einstein's equation gives: \[ K_P=h\nu_P-\phi \] The problem states that P is above threshold, so: \[ h\nu_P\gt \phi \] and: \[ K_P\gt 0 \] \(\textbf{Photoelectron kinetic energy under Q:}\) The frequency of Q is: \[ \nu_Q=2\nu_P \] Therefore: \[ K_Q=h(2\nu_P)-\phi \] \[ K_Q=2h\nu_P-\phi \] \(\textbf{Compare with twice the original kinetic energy:}\) Twice the original maximum kinetic energy is: \[ 2K_P=2(h\nu_P-\phi) \] \[ 2K_P=2h\nu_P-2\phi \] Subtracting: \[ K_Q-2K_P=(2h\nu_P-\phi)-(2h\nu_P-2\phi) \] \[ K_Q-2K_P=\phi \] For a metal with positive work function: \[ \phi\gt 0 \] Hence: \[ K_Q\gt 2K_P \] \(\textbf{Physical interpretation:}\) The equal photon arrival rates and equal quantum efficiencies give equal saturation currents. However, doubling incident frequency doubles photon energy, not the excess energy remaining after electron escape. Because the fixed work function is subtracted in both experiments, the new maximum kinetic energy is greater than twice the original maximum kinetic energy. \(\textbf{Final answer:}\) \[ \boxed{I_{sQ}=I_{sP},\qquad K_Q\gt 2K_P} \]

Explanation: The electroscope initially indicates that the zinc plate possesses excess negative charge. When suitable ultraviolet radiation illuminates the plate, electrons can escape from the zinc surface. This reduces the magnitude of the plate's negative charge and causes the electroscope leaf divergence to decrease. The subsequent visible-light source delivers greater total radiant power, but this does not guarantee that its individual photons possess sufficient energy to liberate electrons. Photon energy is proportional to frequency and inversely proportional to wavelength in vacuum. Visible radiation generally has lower frequency than ultraviolet radiation. If the visible photon energy is below the escape-energy requirement of zinc, ordinary photoelectric emission cannot occur even when the total incident power is large. The greater power may reflect a greater photon arrival rate rather than greater energy per photon. Visible photons certainly possess energy, so the absence of emission does not mean that their energy is zero. Photons are electrically neutral in both visible and ultraviolet radiation. The observations therefore demonstrate the importance of the threshold condition rather than any special charge carried by ultraviolet photons. They also connect Hallwachs' charge observations with the frequency-dependent nature of photoelectric emission.

Explanation: The initially negatively charged zinc plate contains an excess of electrons. Suitable ultraviolet radiation causes some electrons to escape from its surface. Each escaping electron removes negative charge and therefore increases the net charge of the remaining plate. Initially, this makes the plate less negative. Since electroscope leaf divergence depends on the magnitude of the charge, the divergence decreases as the negative charge approaches zero. When sufficient electrons have escaped to neutralise the plate, the leaves can collapse completely. Continued electron emission can then leave the plate positively charged. Further loss of electrons increases the magnitude of this positive charge, causing the leaves to diverge again. Thus, the same process of electron emission can produce decreasing divergence followed by increasing divergence. The electroscope leaf separation alone does not indicate whether the plate is positively or negatively charged. This distinction is essential when interpreting Hallwachs' observations because the direction of charge transfer remains unchanged even though the visible electroscope response reverses.

Explanation: \(\textbf{Effect of electron escape on plate charge:}\) An electron carries negative charge. When electrons escape from a plate, its remaining net charge increases by the magnitude of the negative charge removed. Thus: \[ Q_f=Q_i+\Delta Q \] where \(\Delta Q\) is the positive increase in plate charge produced by electron emission. \(\textbf{Final charge on P:}\) Initially: \[ Q_{iP}=-5.0\,\text{nC} \] The charge increase is: \[ \Delta Q_P=+3.0\,\text{nC} \] Therefore: \[ Q_{fP}=-5.0+3.0 \] \[ Q_{fP}=-2.0\,\text{nC} \] \(\textbf{Final charge on Q:}\) Initially: \[ Q_{iQ}=+5.0\,\text{nC} \] The same electron loss produces: \[ \Delta Q_Q=+3.0\,\text{nC} \] Hence: \[ Q_{fQ}=5.0+3.0 \] \[ Q_{fQ}=+8.0\,\text{nC} \] \(\textbf{Compare electroscope responses:}\) The two electroscopes are identical. For comparable conditions, a greater magnitude of net charge produces greater leaf divergence. The final charge magnitudes are: \[ |Q_{fP}|=2.0\,\text{nC} \] \[ |Q_{fQ}|=8.0\,\text{nC} \] Therefore, Q has the greater final leaf divergence. \(\textbf{Physical interpretation:}\) Both plates lose the same number of electrons, so their signed charge changes are identical. However, their initial charges have opposite signs. Electron loss decreases the magnitude of P's negative charge while increasing the magnitude of Q's positive charge. Equal electron emission therefore need not produce equal changes in electroscope divergence. \(\textbf{Final answer:}\) \[ \boxed{Q_{fP}=-2.0\,\text{nC},\qquad Q_{fQ}=+8.0\,\text{nC}} \]

Explanation: Suitable ultraviolet photons can liberate electrons from either zinc surface because their energies exceed the required escape energy. The fundamental microscopic process is therefore initially the same for the two plates. However, the electrical boundary conditions differ. Plate P is isolated, so electrons leaving it are not automatically replaced through an external conductor. Their removal leaves an increasing positive net charge on P. This accumulated positive charge can eventually oppose further electron escape through electrostatic attraction. Plate Q, in contrast, is connected to the earth through a conducting wire. The earth can supply electrons to compensate for electrons removed by photoelectric emission. Consequently, Q can remain approximately at earth potential rather than accumulating the same positive charge as an isolated plate. Charge may flow through the earthing conductor while illumination and emission continue. Thus, the amount of charge accumulated on a metal is not necessarily equal to the total charge carried away by emitted photoelectrons when an external conducting path exists. The experiment distinguishes photoelectron emission from charge accumulation and shows why electrical connections matter when interpreting Hallwachs-type observations.

Explanation: \(\textbf{Charge transferred during photoelectric emission:}\) Every emitted electron carries charge: \[ q_e=-e \] If \(N\) electrons leave P, the total negative charge removed from P is: \[ q_{\text{removed}}=-Ne \] Removal of negative charge increases the net charge remaining on P. Since P was initially neutral: \[ Q_P=+Ne \] \(\textbf{Charge received by the collector:}\) Every emitted electron reaches Q. Therefore, Q receives the total charge: \[ Q_Q=N(-e) \] \[ Q_Q=-Ne \] \(\textbf{Total charge of the system:}\) Initially, both plates are neutral. Hence: \[ Q_{\text{initial}}=0 \] After the transfer: \[ Q_{\text{final}}=Q_P+Q_Q \] Substituting: \[ Q_{\text{final}}=Ne-Ne \] \[ Q_{\text{final}}=0 \] \(\textbf{Physical interpretation:}\) The ultraviolet photons supply the energy required to liberate electrons from P, but they do not carry the transferred electric charge. The electrons themselves transport negative charge from the emitter to the collector. The emitter becomes positively charged because it loses electrons, while the collector becomes negatively charged because it gains them. No electric charge is created or destroyed. If the plates remain electrically isolated after illumination, the separated charges may persist even though no sustained external circuit current flows. \(\textbf{Final answer:}\) \[ \boxed{Q_P=+Ne,\qquad Q_Q=-Ne,\qquad Q_{\text{total}}=0} \]

Explanation: \(\textbf{Charge lost through electron emission:}\) In the isolated-plate experiment, the increase in plate charge is: \[ \Delta Q=+4.8\,\text{nC} \] The positive charge increase results from the removal of negative electrons. Its magnitude is: \[ |\Delta Q|=4.8\times10^{-9}\,\text{C} \] \(\textbf{Determine the electron emission rate:}\) The pulse duration is: \[ \Delta t=10\,\text{ms} \] \[ \Delta t=10\times10^{-3}\,\text{s} \] The magnitude of charge emitted per second is: \[ I_{\text{emission}}=\frac{|\Delta Q|}{\Delta t} \] Substituting: \[ I_{\text{emission}}=\frac{4.8\times10^{-9}}{10\times10^{-3}}\,\text{A} \] \[ I_{\text{emission}}=4.8\times10^{-7}\,\text{A} \] \[ I_{\text{emission}}=0.48\,\mu\text{A} \] This is the current magnitude that would correspond to complete collection of the emitted electrons. \(\textbf{Use the Lenard-tube measurement:}\) The actual average collected photocurrent is: \[ I_{\text{collected}}=0.36\,\mu\text{A} \] Let \(f\) be the fraction of emitted electrons reaching the collector. Then: \[ I_{\text{collected}}=fI_{\text{emission}} \] Therefore: \[ f=\frac{I_{\text{collected}}}{I_{\text{emission}}} \] Substituting: \[ f=\frac{0.36}{0.48} \] \[ f=0.75 \] Expressed as a percentage: \[ f=75\% \] \(\textbf{Physical interpretation:}\) The isolated-plate experiment measures the total negative charge removed from the emitter. The evacuated-tube experiment measures only the emitted electrons that successfully reach the collector. Because collection is incomplete, the measured photocurrent is smaller than the current corresponding to the total emission rate. The comparison therefore determines collection efficiency without requiring knowledge of photon energy or the work function. \(\textbf{Final answer:}\) \[ \boxed{f=75\%} \]

Explanation: Equal incident radiant powers mean that both sources deliver equal total energies per second to the photosensitive surface. However, the energies carried by individual photons are different. Since each photon from Q carries twice the energy of a photon from P, Q supplies only half as many photons per second as P. This establishes the photon arrival-rate ratio, but it does not by itself establish the photoelectron emission-rate ratio. Photoelectric quantum efficiency describes the fraction of incident photons that successfully produce emitted photoelectrons. The problem does not state that the efficiencies are equal for the two frequencies. Consequently, the two sources may produce different fractions of successful photoelectric events. The saturation current is proportional to the number of emitted electrons collected per second, not simply to the incident photon arrival rate. If the efficiencies were equal, P would produce twice the saturation current of Q. Without that additional condition, however, Q could produce a comparable or even greater current if its quantum efficiency were sufficiently larger. The important experimental lesson is that photon arrival rate, photon energy and photoelectric quantum efficiency must be distinguished before inferring photocurrent from radiant power.

Explanation: Suitable ultraviolet photons initially supply sufficient energy for electrons to escape from the zinc surface. Because the plate is electrically isolated, the departing electrons leave positive net charge behind. As additional electrons escape, the positive charge of the plate increases. This produces an electrostatic field that attracts negatively charged electrons back towards the plate. Consequently, an escaping electron must overcome an increasing electrostatic energy barrier in addition to the microscopic escape requirement of the material. Eventually, the positive charge may become large enough to suppress further successful electron escape into the surroundings. The incident photon frequency has not changed, so this suppression cannot be attributed to decreasing photon energy. Briefly connecting the plate to earth allows charge redistribution that can reduce its accumulated positive potential. After the plate is isolated again, the electrostatic opposition to electron escape is smaller. Ultraviolet illumination can therefore produce appreciable electron emission once more. This changed-condition experiment distinguishes an electrostatic charging effect from a permanent change in the intrinsic work function. It also illustrates why measurements with isolated plates and measurements using continuously connected photoelectric circuits can behave differently.