Dual Nature of Matter and Radiation MCQs with Answers – Part 1 (NEET Physics)

Explanation: The formation of alternate bright and dark fringes is a characteristic consequence of wave interference. When light waves from two coherent slits overlap, their electric fields combine according to the principle of superposition. At positions where the waves arrive with suitable phase agreement, constructive interference produces greater intensity. At positions where their phase difference corresponds to destructive interference, the resultant intensity becomes minimum. The spatial distribution of bright and dark fringes therefore depends on the relative phases of the overlapping waves. A classical wave description successfully accounts for this pattern. Photon energy quantisation, by itself, does not explain the positions of interference maxima and minima. This observation establishes an important aspect of the wave character of electromagnetic radiation without excluding its particle-like behaviour in other interactions.

Explanation: Statement I is correct because diffraction is the spreading of a wave when it passes through an opening or encounters an obstacle. The effect becomes particularly noticeable when the relevant dimensions are comparable to the wavelength. For a much wider aperture, diffraction still occurs, but its angular spreading is relatively small. Thus, diffraction provides evidence for the wave character of light. Statement II is incorrect because polarisation is a characteristic property of transverse waves. Electromagnetic waves have oscillating electric and magnetic fields perpendicular to their direction of propagation. A polariser selects a particular transverse direction of electric-field oscillation. Longitudinal waves cannot exhibit polarisation in this manner. Therefore, diffraction and polarisation both support the wave description of light, while polarisation additionally establishes its transverse character.

Explanation: The assertion is true because photon energy increases as wavelength decreases in vacuum. The reason is also true. For electromagnetic radiation in vacuum, the product of frequency and wavelength equals the constant speed of light. Consequently, a smaller wavelength corresponds to a greater frequency. Planck's photon-energy relation states that the energy of one photon is \(E=h\nu\). Substituting the frequency-wavelength relation gives \(E=hc/\lambda\). Thus, photon energy varies inversely with wavelength. The reason directly establishes the physical relationship stated in the assertion. Importantly, this comparison concerns the energy of an individual photon and does not require knowledge of the intensity or total power of either beam.

Explanation: \(\textbf{Given:}\) The wavelength and physical constants are: \[ \lambda=500\times10^{-9}\,\text{m} \] \[ h=6.63\times10^{-34}\,\text{J s},\qquad c=3.0\times10^8\,\text{m s}^{-1} \] \(\textbf{Photon-energy relation:}\) The energy of a photon is related to its frequency by: \[ E=h\nu \] For electromagnetic radiation travelling through vacuum: \[ c=\nu\lambda \] Therefore: \[ E=\frac{hc}{\lambda} \] \(\textbf{Substitution:}\) \[ E=\frac{(6.63\times10^{-34})(3.0\times10^8)}{500\times10^{-9}}\,\text{J} \] Simplifying: \[ E=\frac{19.89\times10^{-26}}{5.00\times10^{-7}}\,\text{J} \] \[ E=3.978\times10^{-19}\,\text{J} \] \(\textbf{Conversion into electronvolts:}\) \[ E=\frac{3.978\times10^{-19}}{1.60\times10^{-19}}\,\text{eV} \] \[ E=2.48625\,\text{eV} \] The result represents the energy of one photon rather than the total energy of the radiation beam. Its value is consistent with the inverse relationship between photon energy and wavelength. \(\textbf{Final answer:}\) \[ \boxed{E\approx2.49\,\text{eV}} \]

Explanation: The energy of an individual photon is determined by its frequency through \(E=h\nu\). Since beams P and Q have the same frequency, their individual photons possess equal energies. Intensity describes the radiant energy crossing unit area per unit time. Both beams illuminate equal areas for equal durations, so the total incident energy is proportional to their respective intensities. Beam P therefore delivers three times the total energy delivered by beam Q. Because the energy per photon is unchanged, this increased energy must correspond to three times as many photons. The detector's complete absorption ensures that all the incident photon energy is included in the comparison. Increasing intensity at fixed frequency does not increase the energy of each photon. This distinction between photon energy and photon number is fundamental to understanding the particle description of radiation.

Explanation: \(\textbf{Photon energies:}\) For radiation travelling through vacuum: \[ E=\frac{hc}{\lambda} \] Hence, the photon energies of the two beams are: \[ E_P=\frac{hc}{\lambda} \] \[ E_Q=\frac{hc}{2\lambda}=\frac{E_P}{2} \] Thus, every photon in beam Q has half the energy of a photon in beam P. \(\textbf{Power comparison:}\) If \(R\) photons arrive per second, the radiant power is: \[ P=RE \] The two beams have the same photon arrival rate \(R\). Therefore: \[ \frac{P_Q}{P_P}=\frac{RE_Q}{RE_P} \] \[ \frac{P_Q}{P_P}=\frac{1}{2} \] The photon number per second remains unchanged, but the total energy delivered per second decreases because every photon carries less energy. Equal photon arrival rates do not necessarily imply equal radiant powers. \(\textbf{Final answer:}\) \[ \boxed{\frac{E_Q}{E_P}=\frac{1}{2},\qquad\frac{P_Q}{P_P}=\frac{1}{2}} \]

Explanation: An interference pattern is evidence that the spatial behaviour of light has a wave-like character. In the double-slit arrangement, the observed distribution is determined by the interference associated with the two paths. A sensitive detector, however, may register energy deposition through individual localised events. Such detection is consistent with the quantum description of radiation. Reducing the intensity decreases the rate at which detection events accumulate, but does not by itself eliminate the characteristic interference distribution. After a sufficiently large number of events, the spatial pattern can still reveal the wave-like behaviour. Thus, localisation of energy transfer does not require abandonment of the wave description of propagation. Similarly, the observation of interference does not imply that energy must always be absorbed continuously. The two descriptions address different observable aspects of the same radiation rather than two permanently separate categories of light.

Explanation: Experiment P demonstrates diffraction because the light spreads noticeably after passing through a narrow aperture. This behaviour is explained naturally using the wave description of electromagnetic radiation. Experiment Q concerns the photoelectric effect, in which electrons are emitted from a suitable material when radiation of adequate frequency is absorbed. The photon picture explains why energy is delivered in discrete quanta and why the frequency is important. Experiment R demonstrates polarisation because the transmitted intensity depends on the relative orientation of the analyser. Polarisation provides evidence that light is a transverse electromagnetic wave. Although both diffraction and polarisation are wave phenomena, they establish different aspects of wave behaviour. Photoelectric emission highlights the particle-like energy-transfer aspect of radiation. These experiments therefore provide complementary evidence for the dual nature of light. None requires the conclusion that radiation permanently changes from one physical species into another.

Explanation: \(\textbf{Frequency comparison:}\) The given frequencies satisfy: \[ \nu_P=2\nu_Q \] The energy of one photon is: \[ E=h\nu \] Therefore: \[ \frac{E_P}{E_Q}=\frac{\nu_P}{\nu_Q}=2 \] A photon belonging to beam P carries twice the energy of a photon belonging to beam Q. \(\textbf{Intensity and photon flux:}\) Let \(F\) represent the number of photons arriving per unit area per second. Then: \[ I=Fh\nu \] Since the two beams have equal intensities: \[ F_Ph\nu_P=F_Qh\nu_Q \] Cancelling \(h\) and rearranging: \[ \frac{F_P}{F_Q}=\frac{\nu_Q}{\nu_P} \] \[ \frac{F_P}{F_Q}=\frac{1}{2} \] Hence, beam Q supplies twice as many photons per unit area per second. Equal intensities require equal energy delivery rates per unit area, not equal photon arrival rates. \(\textbf{Final answer:}\) \[ \boxed{\frac{E_P}{E_Q}=2,\qquad\frac{F_Q}{F_P}=2} \]

Explanation: Photoelectric emission requires the energy of an individual incident photon to be sufficient to liberate an electron from the metal surface. For the given material, the minimum required photon energy corresponds to the threshold frequency \(\nu_0\). Beam P has a frequency smaller than this threshold, so its individual photons do not possess sufficient energy for ordinary single-photon photoelectric emission. Its intensity is five times greater, but this increases the rate of photon arrival rather than the energy carried by each photon. Consequently, the greater intensity cannot compensate for the insufficient individual photon energy. Beam Q has a frequency greater than the threshold and therefore contains photons capable of producing photoelectric emission. Its lower intensity may affect how many photoelectrons are emitted per second, but it does not prevent emission. The decisive condition is the photon frequency relative to the material's threshold. This observation is one reason why a simple classical continuous-energy-absorption picture cannot account for the photoelectric effect.

Explanation: \(\textbf{Photon energies:}\) For a photon of wavelength \(\lambda\): \[ E_1=\frac{hc}{\lambda} \] For a photon of wavelength \(3\lambda\): \[ E_2=\frac{hc}{3\lambda} \] Therefore: \[ E_2=\frac{E_1}{3} \] \(\textbf{Equal-power condition:}\) Let both components deliver the same power \(P\). If \(R_1\) and \(R_2\) are their respective photon arrival rates: \[ P=R_1E_1=R_2E_2 \] Hence: \[ \frac{R_2}{R_1}=\frac{E_1}{E_2}=3 \] The longer-wavelength component supplies three times as many photons per second. \(\textbf{Fraction of total photons:}\) The required fraction is: \[ f=\frac{R_2}{R_1+R_2} \] Substituting \(R_2=3R_1\): \[ f=\frac{3R_1}{R_1+3R_1} \] \[ f=\frac{3}{4} \] The two components contribute equal energy per second but unequal photon numbers. The lower-energy photons must be more numerous to maintain the same power. \(\textbf{Final answer:}\) \[ \boxed{f=\frac{3}{4}} \]

Explanation: \(\textbf{Photon flux relation:}\) Photon arrival rate per unit area is obtained by dividing intensity by photon energy: \[ F=\frac{I}{E} \] For uniform illumination: \[ I=\frac{P}{A} \] Photon energy is: \[ E=\frac{hc}{\lambda} \] Combining these relations: \[ F=\frac{P\lambda}{Ahc} \] \(\textbf{Compare the two beams:}\) For the first beam: \[ F_1=\frac{P(400\,\text{nm})}{Ahc} \] For the second beam: \[ F_2=\frac{2P(600\,\text{nm})}{3Ahc} \] Taking the ratio: \[ \frac{F_2}{F_1}=\frac{2P}{P}\times\frac{600}{400}\times\frac{A}{3A} \] \[ \frac{F_2}{F_1}=2\times\frac{3}{2}\times\frac{1}{3} \] \[ \frac{F_2}{F_1}=1 \] \(\textbf{Physical interpretation:}\) The second beam has twice the total power, but its photons have smaller individual energies because their wavelength is longer. These changes increase the total photon arrival rate by a factor of three. However, the illuminated area also increases by a factor of three. Consequently, the number of photons incident per second on each unit area remains unchanged. \(\textbf{Final answer:}\) \[ \boxed{\frac{F_2}{F_1}=1} \]

Explanation: \(\textbf{Relate photon energy to radiant power:}\) For a monochromatic source emitting \(R\) photons per second: \[ P=RE \] Therefore, the energy of each emitted photon is: \[ E=\frac{P}{R} \] \(\textbf{Calculate photon energy for source P:}\) \[ E_P=\frac{3.0\times10^{-3}}{3.0\times10^{15}}\,\text{J} \] \[ E_P=1.0\times10^{-18}\,\text{J} \] \(\textbf{Calculate photon energy for source Q:}\) \[ E_Q=\frac{2.0\times10^{-3}}{4.0\times10^{15}}\,\text{J} \] \[ E_Q=5.0\times10^{-19}\,\text{J} \] \(\textbf{Relate energy to wavelength:}\) For electromagnetic radiation in vacuum: \[ E=\frac{hc}{\lambda} \] Therefore: \[ \lambda\propto\frac{1}{E} \] Taking the wavelength ratio: \[ \frac{\lambda_P}{\lambda_Q}=\frac{E_Q}{E_P} \] Substituting: \[ \frac{\lambda_P}{\lambda_Q}=\frac{5.0\times10^{-19}}{1.0\times10^{-18}} \] \[ \frac{\lambda_P}{\lambda_Q}=\frac{1}{2} \] \(\textbf{Physical interpretation:}\) Source P emits photons having twice the energy of those emitted by Q. Since photon energy is inversely proportional to vacuum wavelength, the wavelength emitted by P is half that emitted by Q. Comparing source powers alone would be insufficient because the sources have different photon emission rates. \(\textbf{Final answer:}\) \[ \boxed{\frac{\lambda_P}{\lambda_Q}=\frac{1}{2}} \]

Explanation: \(\textbf{Detected and incident photon rates:}\) The detected photon rate is: \[ R_d=2.0\times10^{15}\,\text{s}^{-1} \] Only \(20\%\) of the incident photons are detected. Thus: \[ R_d=0.20R_i \] where \(R_i\) is the incident photon rate. Rearranging: \[ R_i=\frac{R_d}{0.20} \] \[ R_i=\frac{2.0\times10^{15}}{0.20} \] \[ R_i=1.0\times10^{16}\,\text{s}^{-1} \] \(\textbf{Energy of an incident photon:}\) \[ E=\frac{hc}{\lambda} \] Substituting the given constants: \[ E=\frac{(6.63\times10^{-34})(3.0\times10^8)}{500\times10^{-9}}\,\text{J} \] \[ E=3.978\times10^{-19}\,\text{J} \] \(\textbf{Incident radiant power:}\) The incident power is the product of incident photon rate and energy per photon: \[ P_i=R_iE \] \[ P_i=(1.0\times10^{16})(3.978\times10^{-19})\,\text{W} \] \[ P_i=3.978\times10^{-3}\,\text{W} \] \[ P_i\approx4.0\,\text{mW} \] \(\textbf{Physical check:}\) The detector misses \(80\%\) of the incident photons, so using only the detected rate would underestimate incident power by a factor of five. The calculated incident power is therefore greater than the power represented by the detected photons. \(\textbf{Final answer:}\) \[ \boxed{P_i\approx4.0\,\text{mW}} \]

Explanation: \(\textbf{Intensity after polarisation:}\) Let the incident intensity be \(I\). For the specified unpolarised beam and ideal polariser: \[ I_t=\frac{I}{2} \] The original detector has collecting area \(A\). Therefore, the incident power intercepted by the original detector is: \[ P_1=IA \] \(\textbf{Power collected after the change:}\) The new detector has an area \(2A\), and the transmitted beam has intensity \(I/2\). Hence: \[ P_2=I_t(2A) \] \[ P_2=\frac{I}{2}(2A) \] \[ P_2=IA=P_1 \] \(\textbf{Photon counting:}\) The frequency of radiation is not changed by an ideal polariser. Thus, the energy of every transmitted photon remains: \[ E=h\nu \] Photon arrival rate equals collected power divided by energy per photon: \[ R_2=\frac{P_2}{h\nu} \] Using \(P_2=P_1=Rh\nu\): \[ R_2=\frac{Rh\nu}{h\nu} \] \[ R_2=R \] \(\textbf{Physical interpretation:}\) The polariser reduces photon flux per unit area by half without changing photon energy. Doubling the detector area compensates for this reduction in flux density. The second counter therefore records the original photon rate. This situation connects the wave-based control of light intensity through polarisation with the photon-based description of energy transfer. \(\textbf{Final answer:}\) \[ \boxed{R_2=R,\qquad E=h\nu} \]

Explanation: \(\textbf{Meaning of graph slope:}\) For monochromatic radiation, the radiant power is: \[ P_{\mathrm{rad}}=RE \] Here, \(R\) denotes the number of photons emitted per second and \(E\) denotes the energy of one photon. Therefore, the slope of a power-versus-photon-rate graph is: \[ m=\frac{P_{\mathrm{rad}}}{R} \] \[ m=E \] Thus, the graph slope directly represents the energy of an individual photon. \(\textbf{Compare photon energies:}\) The given slope relationship is: \[ m_P=3m_Q \] Consequently: \[ E_P=3E_Q \] \(\textbf{Convert energy ratio into wavelength ratio:}\) For electromagnetic radiation in vacuum: \[ E=\frac{hc}{\lambda} \] Substituting into the energy relationship: \[ \frac{hc}{\lambda_P}=3\frac{hc}{\lambda_Q} \] Cancelling the common factor \(hc\): \[ \frac{1}{\lambda_P}=\frac{3}{\lambda_Q} \] Rearranging: \[ \lambda_Q=3\lambda_P \] \(\textbf{Substitution:}\) \[ \lambda_Q=3(300\,\text{nm}) \] \[ \lambda_Q=900\,\text{nm} \] \(\textbf{Physical check:}\) The steeper graph corresponds to photons with greater energy. Since source P has three times the photon energy of source Q, its wavelength must be three times smaller. The inferred wavelength of Q therefore agrees with the inverse energy-wavelength relation. \(\textbf{Final answer:}\) \[ \boxed{\lambda_Q=900\,\text{nm}} \]

Explanation: The positions of interference fringes are determined by the wavelength, slit geometry and relative phase of the interfering waves. The neutral-density filter does not change these quantities. Consequently, the positions and spacing of the interference fringes remain unchanged. The filter does, however, reduce the energy arriving per unit area per unit time. Since the frequency is unchanged, the energy of each photon remains \(E=h\nu\). The reduced intensity therefore corresponds to a smaller photon arrival rate. Under the stated uniform attenuation, the expected photon detections at each position decrease to one-fourth of their original values. The relative distribution of detections across the screen remains the same. In particular, the interference pattern is not compressed merely because fewer photons arrive. This illustrates how a wave-like spatial distribution can coexist with discrete photon detection events and how intensity can change independently of photon energy.

Explanation: \(\textbf{Frequency at the boundary:}\) The frequency of electromagnetic radiation is determined by its source and remains unchanged when the radiation enters a stationary transparent medium. Let the frequency be \(\nu\). In vacuum: \[ c=\nu\lambda_0 \] The given vacuum wavelength is: \[ \lambda_0=600\,\text{nm} \] \(\textbf{Wavelength inside the medium:}\) The propagation speed inside the medium is: \[ v=\frac{2c}{3} \] Using the wave relation in the medium: \[ v=\nu\lambda_m \] Dividing the medium relation by the vacuum relation: \[ \frac{\lambda_m}{\lambda_0}=\frac{v}{c} \] Substituting: \[ \frac{\lambda_m}{600\,\text{nm}}=\frac{2}{3} \] \[ \lambda_m=\frac{2}{3}\times600\,\text{nm} \] \[ \lambda_m=400\,\text{nm} \] \(\textbf{Photon-energy comparison:}\) Photon energy depends on frequency: \[ E=h\nu \] Since frequency is unchanged: \[ E_m=E_0 \] The reduction in wavelength is caused by the reduction in propagation speed, not by an increase in frequency. The vacuum expression \(E=hc/\lambda_0\) must not be used with the wavelength inside the medium while retaining the vacuum speed \(c\). \(\textbf{Final answer:}\) \[ \boxed{\lambda_m=400\,\text{nm},\qquad E_m=E_0} \]

Explanation: \(\textbf{Convert total power into energy per second:}\) The emitted power is: \[ P=0.64\times10^{-3}\,\text{J s}^{-1} \] Converting to electronvolts per second: \[ P=\frac{0.64\times10^{-3}}{1.60\times10^{-19}}\,\text{eV s}^{-1} \] \[ P=4.0\times10^{15}\,\text{eV s}^{-1} \] \(\textbf{Average energy per emitted photon:}\) The total photon emission rate is: \[ R=1.0\times10^{15}\,\text{s}^{-1} \] Therefore: \[ E_{\text{avg}}=\frac{P}{R} \] \[ E_{\text{avg}}=\frac{4.0\times10^{15}}{1.0\times10^{15}}\,\text{eV} \] \[ E_{\text{avg}}=4.0\,\text{eV} \] \(\textbf{Resolve the two photon populations:}\) Let \(f\) be the fraction of photons having energy \(5\,\text{eV}\). Then the remaining fraction \(1-f\) consists of photons having energy \(2\,\text{eV}\). The average photon energy is: \[ E_{\text{avg}}=f(5)+(1-f)(2) \] Substituting the measured average: \[ 4=5f+2-2f \] \[ 4=2+3f \] \[ 3f=2 \] \[ f=\frac{2}{3} \] \(\textbf{Consistency check:}\) The higher-energy photons contribute more energy per photon, so their contribution to total radiant power is not the same as their numerical fraction. Substitution gives the required average energy of \(4\,\text{eV}\), which lies between the two allowed photon energies. \(\textbf{Final answer:}\) \[ \boxed{f=\frac{2}{3}} \]

Explanation: The first observation demonstrates that a large incident intensity is not sufficient to guarantee photoelectric emission. The second demonstrates that emission can occur even when the incident intensity is extremely small. The essential distinction is the frequency of the incident radiation. According to the photon description, the energy of an individual photon is proportional to frequency. A photon must supply sufficient energy to enable an electron to escape the metal surface. If its energy is below the required threshold, increasing the number of such photons does not produce ordinary single-photon photoelectric emission. Conversely, sufficiently energetic photons can liberate electrons even when their arrival rate is small. The absence of a measurable delay at low intensity is consistent with energy being transferred during individual photon-electron interactions rather than gradually accumulated from a continuous electromagnetic wave. The experiment therefore supports the distinction between photon energy and incident intensity. It does not imply that frequency is controlled by intensity or that a greater intensity automatically produces greater energy per photon.