Atomic Structure MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: Wavelength, frequency, time period and wavenumber describe different characteristics of electromagnetic radiation. Wavelength \(\lambda\) is the distance between two successive points of a wave that are in the same phase, such as two consecutive crests. Frequency \(\nu\) is the number of complete oscillations passing a fixed point per second. Its SI unit is hertz, equivalent to \(\mathrm{s^{-1}}\). The time period \(T\) is the time required for one complete oscillation, so frequency and time period are reciprocals. Wavenumber \(\tilde{\nu}\) represents the number of wavelengths per unit length and is the reciprocal of wavelength. The SI unit of wavenumber is \(\mathrm{m^{-1}}\), although \(\mathrm{cm^{-1}}\) is commonly used in spectroscopy. Frequency and wavenumber are therefore not interchangeable: frequency counts oscillations per unit time, whereas wavenumber counts wavelengths per unit distance. For electromagnetic radiation travelling in vacuum, they are connected through \(\nu=c\tilde{\nu}\), where \(c\) is the speed of light.

Explanation: \(\textbf{Frequency comparison:}\) Both radiations travel through vacuum, so their propagation speed is the same. The governing wave relation is \[ c=\nu\lambda \] For the two radiations, \[ \nu_A\lambda_A=\nu_B\lambda_B \] The supplied frequency condition is \[ \nu_B=\frac{3}{2}\nu_A \] Substituting into the wave relation, \[ \lambda_B=\lambda_A\frac{\nu_A}{\nu_B} \] \[ \lambda_B=500\times\frac{2}{3}=333.3\,\mathrm{nm} \] \(\textbf{Time-period comparison:}\) Frequency and time period satisfy \[ T=\frac{1}{\nu} \] Therefore, \[ \frac{T_B}{T_A}=\frac{\nu_A}{\nu_B}=\frac{2}{3} \] \(\textbf{Physical consistency:}\) Radiation B has a higher frequency, so it must have both a shorter wavelength and a shorter time period. The wavelength and time-period ratios are identical because both quantities vary inversely with frequency when the speed is fixed. A ratio greater than unity for either quantity would contradict the given increase in frequency. \(\textbf{Final answer:}\) \[ \boxed{\lambda_B=333.3\,\mathrm{nm},\qquad \frac{T_B}{T_A}=\frac{2}{3}} \]

Explanation: \(\textbf{Relations used for verification:}\) For every radiation travelling through vacuum, \[ \nu=\frac{c}{\lambda} \] The corresponding wavenumber is \[ \tilde{\nu}=\frac{1}{\lambda} \] Both relations must be satisfied simultaneously by each row. \(\textbf{Check radiation P:}\) \[ \nu_P=\frac{3.0\times10^8}{400\times10^{-9}}=7.50\times10^{14}\,\mathrm{Hz} \] \[ \tilde{\nu}_P=\frac{1}{400\times10^{-9}}=2.50\times10^6\,\mathrm{m^{-1}} \] Both entries are correct. \(\textbf{Check radiation Q:}\) For \(500\,\mathrm{nm}\), the corresponding values are \(6.00\times10^{14}\,\mathrm{Hz}\) and \(2.00\times10^6\,\mathrm{m^{-1}}\). Thus, radiation Q is consistent. \(\textbf{Check radiation R:}\) For \(600\,\mathrm{nm}\), \[ \nu_R=5.00\times10^{14}\,\mathrm{Hz} \] \[ \tilde{\nu}_R=1.667\times10^6\,\mathrm{m^{-1}} \] The displayed wavenumber is correct to three significant figures. \(\textbf{Check radiation S:}\) For \(750\,\mathrm{nm}\), the frequency is \[ \nu_S=\frac{3.0\times10^8}{750\times10^{-9}}=4.00\times10^{14}\,\mathrm{Hz} \] However, its wavenumber should be \[ \tilde{\nu}_S=\frac{1}{750\times10^{-9}}=1.333\times10^6\,\mathrm{m^{-1}} \] The table instead lists \(1.50\times10^6\,\mathrm{m^{-1}}\). Therefore, radiation S has a correct frequency but an incorrect wavenumber. A row may satisfy the frequency relation yet fail the reciprocal-wavelength check, so both quantities should be verified independently.

Explanation: \(\textbf{Interpret the graph:}\) The horizontal axis represents reciprocal wavelength, while the vertical axis represents frequency. Their fundamental relationship is \[ \nu=c\tilde{\nu} \] Therefore, the slope of a frequency-versus-wavenumber graph represents the propagation speed of radiation. Since the graph passes through the origin, frequency is directly proportional to wavenumber. \(\textbf{Determine the slope:}\) \[ \text{Slope}=\frac{\Delta\nu}{\Delta\tilde{\nu}} \] Using the supplied point and the origin, \[ \text{Slope}=\frac{6.0\times10^{14}\,\mathrm{s^{-1}}}{2.0\times10^6\,\mathrm{m^{-1}}} \] \[ c=3.0\times10^8\,\mathrm{m\,s^{-1}} \] The dimensions also confirm the interpretation because \(\mathrm{s^{-1}/m^{-1}}=\mathrm{m\,s^{-1}}\). \(\textbf{Find the new frequency:}\) At the specified wavenumber, \[ \nu=(3.0\times10^8)(3.0\times10^6) \] \[ \nu=9.0\times10^{14}\,\mathrm{Hz} \] \(\textbf{Convert frequency into time period:}\) \[ T=\frac{1}{\nu} \] \[ T=\frac{1}{9.0\times10^{14}}=1.11\times10^{-15}\,\mathrm{s} \] \(\textbf{Consistency check:}\) The new wavenumber is \(1.5\) times the wavenumber of the measured point, so its frequency must also be \(1.5\) times as large. Its time period must consequently be two-thirds of the original time period. This confirms the direction of change without requiring another calculation. \(\textbf{Final answer:}\) \[ \boxed{T=1.11\times10^{-15}\,\mathrm{s}} \]

Explanation: Statement I is correct because electromagnetic radiation consists of oscillating electric and magnetic fields. In the conventional electromagnetic-wave description, these fields oscillate perpendicular to each other and to the direction in which the wave propagates. Unlike mechanical waves such as sound, electromagnetic waves do not require a material medium for propagation. Statement II is correct because ultraviolet radiation lies beyond the violet end of the visible spectrum in the direction of shorter wavelengths. For radiation in vacuum, frequency and wavelength are inversely related through \(c=\nu\lambda\). Consequently, ultraviolet radiation has a higher frequency than visible violet light. Statement III is incorrect because red light has a longer wavelength and lower frequency than violet light. Since \(T=1/\nu\), the lower frequency of red light corresponds to a longer, not shorter, time period. Statement IV is also incorrect because all electromagnetic radiations travel through vacuum with the same speed \(c\). Radio waves, visible light, ultraviolet radiation and X-rays differ in frequency and wavelength, but not in their vacuum propagation speed. The decisive distinction is between propagation speed, which remains constant in vacuum, and the frequency-wavelength combination, which varies across the electromagnetic spectrum. Therefore, Statements I and II are the only correct statements.

Explanation: The Assertion is true because the energy carried by an individual photon depends on its frequency, whereas the total energy of a beam also depends on how many photons are present. According to Planck's quantum relation, \[ E_{\mathrm{photon}}=h\nu \] A photon belonging to the beam of frequency \(2\nu\) has twice the energy of a photon belonging to the beam of frequency \(\nu\). However, this does not imply that the higher-frequency beam must carry twice the total energy. For example, a beam containing \(2N\) photons of frequency \(\nu\) carries the same total energy as a beam containing \(N\) photons of frequency \(2\nu\). The Reason is also true because the total energy of \(N\) identical photons is \(Nh\nu\). This relation connects photon count with the energy determined by frequency. It directly explains how two beams with different individual photon energies can nevertheless have equal total energies. A frequent misconception is to use \(E=h\nu\) for the entire beam without accounting for the number of photons. That expression gives the energy of one photon, not automatically the energy of a macroscopic collection of photons. Therefore, both statements are correct, and the Reason provides the explanation for the Assertion.

Explanation: \(\textbf{Photon-energy relation:}\) The energy of one photon is determined by its frequency or wavelength: \[ E=h\nu=\frac{hc}{\lambda} \] The supplied value of \(hc\) is expressed in \(\mathrm{eV\,nm}\), so the wavelength can be inserted directly in nanometres to obtain energy in electronvolts. \(\textbf{Calculate photon energy:}\) \[ E=\frac{1240\,\mathrm{eV\,nm}}{620\,\mathrm{nm}} \] \[ E=2.0\,\mathrm{eV} \] \(\textbf{Calculate frequency independently:}\) For electromagnetic radiation travelling in vacuum, \[ \nu=\frac{c}{\lambda} \] First express the wavelength in SI units: \[ \lambda=620\times10^{-9}\,\mathrm{m} \] Substitute the speed of light: \[ \nu=\frac{3.0\times10^8}{620\times10^{-9}} \] \[ \nu=4.84\times10^{14}\,\mathrm{Hz} \] \(\textbf{Cross-check the two results:}\) The wavelength belongs to the visible region, and the calculated frequency is consistent with that region. A photon energy of approximately \(2\,\mathrm{eV}\) is also consistent with visible red radiation. The two results are connected through the same physical relation, so they must not be selected independently from incompatible option values. \(\textbf{Final answer:}\) \[ \boxed{E=2.0\,\mathrm{eV},\qquad \nu=4.84\times10^{14}\,\mathrm{Hz}} \]

Explanation: \(\textbf{Determine the total emitted energy:}\) Power represents energy emitted per unit time. Therefore, the energy emitted during a specified interval is \[ E_{\mathrm{total}}=Pt \] Convert the supplied quantities into SI units: \[ P=9.94\times10^{-3}\,\mathrm{J\,s^{-1}} \] \[ t=2.00\times10^{-3}\,\mathrm{s} \] Substitution gives \[ E_{\mathrm{total}}=(9.94\times10^{-3})(2.00\times10^{-3}) \] \[ E_{\mathrm{total}}=1.988\times10^{-5}\,\mathrm{J} \] \(\textbf{Find the energy of one photon:}\) Every photon in this monochromatic radiation has the same energy: \[ E_{\mathrm{photon}}=\frac{hc}{\lambda} \] The wavelength is \[ \lambda=500\times10^{-9}\,\mathrm{m} \] Hence, \[ E_{\mathrm{photon}}=\frac{(6.626\times10^{-34})(3.0\times10^8)}{500\times10^{-9}} \] \[ E_{\mathrm{photon}}=3.976\times10^{-19}\,\mathrm{J} \] \(\textbf{Calculate the photon count:}\) The number of photons equals the total energy divided by the energy per photon: \[ N=\frac{E_{\mathrm{total}}}{E_{\mathrm{photon}}} \] \[ N=\frac{1.988\times10^{-5}}{3.976\times10^{-19}} \] \[ N\approx5.0\times10^{13} \] \(\textbf{Physical check:}\) A milliwatt-scale source operating for a millisecond releases energy of the order of \(10^{-5}\,\mathrm{J}\). Each visible photon carries energy of the order of \(10^{-19}\,\mathrm{J}\). Therefore, a photon count of the order of \(10^{13}\) to \(10^{14}\) is reasonable. Using power directly as total energy, without multiplying by the emission time, would produce an incorrect photon count. \(\textbf{Final answer:}\) \[ \boxed{N\approx5.0\times10^{13}\ \text{photons}} \]

Explanation: \(\textbf{Identify the two changing factors:}\) The sources differ in both radiant power and wavelength. The larger power of source B means that it emits more total energy per second. Its longer wavelength means that each of its photons carries less energy. Both effects must be included when comparing the photon emission rates. \(\textbf{Relate power to photon rate:}\) For monochromatic radiation, \[ E_{\mathrm{photon}}=\frac{hc}{\lambda} \] The photon emission rate is \[ \dot N=\frac{P}{E_{\mathrm{photon}}} \] Substituting photon energy, \[ \dot N=\frac{P\lambda}{hc} \] \(\textbf{Compare the two sources:}\) Since \(h\) and \(c\) are common, \[ \frac{\dot N_B}{\dot N_A}=\frac{P_B\lambda_B}{P_A\lambda_A} \] The supplied power ratio is \[ \frac{P_B}{P_A}=\frac{3}{2} \] The wavelength ratio is \[ \frac{\lambda_B}{\lambda_A}=\frac{600}{400}=\frac{3}{2} \] Therefore, \[ \frac{\dot N_B}{\dot N_A}=\frac{3}{2}\times\frac{3}{2} \] \[ \frac{\dot N_B}{\dot N_A}=\frac{9}{4} \] \(\textbf{Interpret the result:}\) Increasing radiant power by a factor of \(1.5\) increases the number of photons supplied per second at a fixed wavelength. Increasing wavelength by another factor of \(1.5\) decreases the energy per photon and therefore permits another factor of \(1.5\) increase in photon rate. The combined result is \(2.25\), not merely \(1.5\). \(\textbf{Final answer:}\) \[ \boxed{\frac{\dot N_B}{\dot N_A}=\frac{9}{4}} \]

Explanation: \(\textbf{Determine the required equivalence:}\) Both sources emit the same total number of photons and deliver the same total energy. Therefore, their average energy per photon must be equal. The first source contains photons of two different wavelengths, so their individual energies must be calculated separately. \(\textbf{Energy of the mixed radiation:}\) For each photon, \[ E=\frac{hc}{\lambda} \] The energy contributed by the \(400\,\mathrm{nm}\) photons is \[ E_1=N\frac{hc}{400\,\mathrm{nm}} \] The energy contributed by the \(800\,\mathrm{nm}\) photons is \[ E_2=3N\frac{hc}{800\,\mathrm{nm}} \] Thus, \[ E_{\mathrm{total}}=Nhc\left(\frac{1}{400}+\frac{3}{800}\right)\mathrm{nm^{-1}} \] \(\textbf{Apply the equal-photon-count condition:}\) The monochromatic source emits \(4N\) photons of an unknown wavelength \(\lambda\). Its total energy is \[ E'_{\mathrm{total}}=\frac{4Nhc}{\lambda} \] Equating the two total energies gives \[ \frac{4Nhc}{\lambda}=Nhc\left(\frac{1}{400}+\frac{3}{800}\right)\mathrm{nm^{-1}} \] Cancel the common factor \(Nhc\): \[ \frac{4}{\lambda}=\left(\frac{1}{400}+\frac{3}{800}\right)\mathrm{nm^{-1}} \] \[ \frac{4}{\lambda}=\frac{2+3}{800}\,\mathrm{nm^{-1}} \] \[ \frac{4}{\lambda}=\frac{5}{800}\,\mathrm{nm^{-1}} \] Consequently, \[ \lambda=\frac{4\times800}{5}\,\mathrm{nm} \] \[ \lambda=640\,\mathrm{nm} \] \(\textbf{Interpretation and consistency:}\) The equivalent wavelength lies between \(400\,\mathrm{nm}\) and \(800\,\mathrm{nm}\), as expected. It is closer to \(800\,\mathrm{nm}\) because three-fourths of the photons in the mixture have that wavelength. However, it is not obtained by taking the arithmetic average of the wavelengths. Photon energies vary inversely with wavelength, so the correct averaging must be performed using reciprocal wavelengths. \(\textbf{Final answer:}\) \[ \boxed{\lambda=640\,\mathrm{nm}} \]

Explanation: \(\textbf{Energy of one photon:}\) The energy carried by an individual photon depends on its wavelength: \[ E_{\mathrm{photon}}=\frac{hc}{\lambda} \] For \(500\,\mathrm{nm}\), \[ \lambda=500\times10^{-9}\,\mathrm{m} \] Substituting, \[ E_{\mathrm{photon}}=\frac{(6.626\times10^{-34})(3.0\times10^8)}{500\times10^{-9}} \] \[ E_{\mathrm{photon}}=3.9756\times10^{-19}\,\mathrm{J} \] \(\textbf{Convert to energy per mole of photons:}\) One mole of photons contains Avogadro's number of photons. Therefore, \[ E_{\mathrm{molar}}=N_AE_{\mathrm{photon}} \] \[ E_{\mathrm{molar}}=(6.022\times10^{23})(3.9756\times10^{-19}) \] \[ E_{\mathrm{molar}}=2.3941\times10^5\,\mathrm{J\,mol^{-1}} \] \[ E_{\mathrm{molar}}=239.41\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Apply the supplied amount:}\) The sample contains only \(0.250\) mole of photons. Hence, \[ E_{\mathrm{total}}=nE_{\mathrm{molar}} \] \[ E_{\mathrm{total}}=0.250\times239.41 \] \[ E_{\mathrm{total}}=59.85\,\mathrm{kJ} \] \(\textbf{Reasonableness check:}\) The calculated energy is one-fourth of the energy corresponding to one mole of photons of the same wavelength. It is important to distinguish the energy of one photon from the energy of a mole of photons. Avogadro's constant converts the microscopic photon energy into a molar quantity; the stated \(0.250\,\mathrm{mol}\) must then be applied. The answer \(239.4\,\mathrm{kJ}\) would correspond to one full mole rather than the quantity supplied. \(\textbf{Final answer:}\) \[ \boxed{E_{\mathrm{total}}\approx59.9\,\mathrm{kJ}} \]

Explanation: \(\textbf{Recognise the non-linear relationship:}\) Photon energy varies inversely with wavelength: \[ E=\frac{hc}{\lambda} \] Therefore, equal increases in wavelength do not necessarily produce equal decreases in photon energy. The energy change must be calculated from the reciprocal wavelengths rather than from the wavelength difference alone. \(\textbf{Case I:}\) The positive decrease in photon energy is \[ \Delta E_1=hc\left(\frac{1}{400}-\frac{1}{500}\right)\mathrm{nm^{-1}} \] Simplifying the bracket, \[ \frac{1}{400}-\frac{1}{500}=\frac{500-400}{400\times500} \] \[ \Delta E_1=hc\left(\frac{100}{200000}\right)\mathrm{nm^{-1}} \] \(\textbf{Case II:}\) Similarly, \[ \Delta E_2=hc\left(\frac{1}{600}-\frac{1}{700}\right)\mathrm{nm^{-1}} \] \[ \Delta E_2=hc\left(\frac{700-600}{600\times700}\right)\mathrm{nm^{-1}} \] \[ \Delta E_2=hc\left(\frac{100}{420000}\right)\mathrm{nm^{-1}} \] \(\textbf{Compare the magnitudes:}\) The factors \(h\) and \(c\), the common wavelength units and the numerator \(100\) cancel in the ratio. Therefore, \[ \frac{\Delta E_1}{\Delta E_2}=\frac{100/200000}{100/420000} \] \[ \frac{\Delta E_1}{\Delta E_2}=\frac{420000}{200000} \] \[ \frac{\Delta E_1}{\Delta E_2}=2.10 \] \(\textbf{Interpretation:}\) Both cases involve a wavelength increase of \(100\,\mathrm{nm}\). However, the change occurring at the shorter-wavelength end produces the larger photon-energy difference. This happens because the magnitude of the energy change over a fixed wavelength interval becomes smaller as the wavelength interval shifts toward longer wavelengths. The ratio must therefore exceed unity, providing a useful check on the result. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Delta E_1}{\Delta E_2}=\frac{21}{10}} \]

Explanation: The Assertion is correct because photoelectric emission requires an individual incident photon to supply at least the minimum energy needed to remove an electron from the metal. This minimum energy is called the work function, represented by \(\phi\). The corresponding minimum incident frequency is the threshold frequency \(\nu_0\), satisfying \(\phi=h\nu_0\). For incident radiation with \(\nu\lt \nu_0\), the energy \(h\nu\) of an individual photon is smaller than the work function. An electron therefore cannot receive sufficient energy from a single photon to escape from the surface in the ordinary photoelectric-effect description. The Reason is also correct because increasing intensity at fixed frequency increases the total radiant energy supplied per unit time by increasing the photon rate. It does not alter the energy \(h\nu\) carried by each photon. Thus, even a much larger number of photons cannot compensate for the insufficient energy of an individual below-threshold photon. When the frequency exceeds the threshold, increasing the intensity can increase the number of photoelectrons emitted under otherwise comparable conditions. However, it does not increase their maximum kinetic energy at fixed frequency. The Reason directly explains the limitation described in the Assertion, so both are true and the Reason is the correct explanation.

Explanation: \(\textbf{Determine the threshold wavelength:}\) The work function is the minimum energy required to remove an electron from the metal surface. It is related to the threshold wavelength by \[ \phi=\frac{hc}{\lambda_0} \] Rearranging, \[ \lambda_0=\frac{hc}{\phi} \] Substituting the given quantities, \[ \lambda_0=\frac{1240\,\mathrm{eV\,nm}}{2.0\,\mathrm{eV}} \] \[ \lambda_0=620\,\mathrm{nm} \] \(\textbf{Check whether emission is possible:}\) The incident wavelength is \(500\,\mathrm{nm}\), which is shorter than the threshold wavelength. A shorter wavelength corresponds to greater photon energy. Therefore, the incident photons possess sufficient energy to eject electrons. \(\textbf{Calculate the incident photon energy:}\) \[ E_{\mathrm{photon}}=\frac{hc}{\lambda} \] \[ E_{\mathrm{photon}}=\frac{1240}{500}=2.48\,\mathrm{eV} \] \(\textbf{Apply Einstein's photoelectric equation:}\) The incident photon energy is divided between overcoming the work function and providing kinetic energy to the electron. \[ K_{\max}=h\nu-\phi \] \[ K_{\max}=2.48-2.0 \] \[ K_{\max}=0.48\,\mathrm{eV} \] \(\textbf{Energy check:}\) The maximum kinetic energy is positive, confirming the possibility of photoelectric emission. It is smaller than the incident photon energy because \(2.0\,\mathrm{eV}\) is required to overcome the work function. The threshold wavelength is determined by the metal, whereas the maximum kinetic energy depends on both the metal and the incident radiation. Confusing these two roles produces incorrect combinations of the threshold wavelength and kinetic energy. \(\textbf{Final answer:}\) \[ \boxed{\lambda_0=620\,\mathrm{nm},\qquad K_{\max}=0.48\,\mathrm{eV}} \]

Explanation: \(\textbf{Compare threshold conditions first:}\) Photoelectric emission is possible when the incident photon energy is at least equal to the work function. In wavelength form, the condition is that the incident wavelength must not exceed the threshold wavelength. The incident radiation has wavelength \(500\,\mathrm{nm}\). For metal P, the threshold wavelength is \(450\,\mathrm{nm}\). Since the incident wavelength is longer than this threshold, its photons have insufficient energy to eject electrons from P. For metal Q, the threshold wavelength is \(650\,\mathrm{nm}\). Since \(500\,\mathrm{nm}\) is shorter than \(650\,\mathrm{nm}\), photoelectric emission occurs from Q. \(\textbf{Calculate the photon energy:}\) \[ E_{\mathrm{photon}}=\frac{1240}{500}=2.48\,\mathrm{eV} \] \(\textbf{Determine the work function of Q:}\) \[ \phi_Q=\frac{hc}{\lambda_{0,Q}} \] \[ \phi_Q=\frac{1240}{650}=1.908\,\mathrm{eV} \] \(\textbf{Calculate the maximum kinetic energy:}\) \[ K_{\max,Q}=E_{\mathrm{photon}}-\phi_Q \] \[ K_{\max,Q}=2.48-1.908 \] \[ K_{\max,Q}=0.572\,\mathrm{eV} \] Thus, the maximum kinetic energy is approximately \(0.57\,\mathrm{eV}\). \(\textbf{Consistency check:}\) Metal P has the shorter threshold wavelength, which corresponds to the larger work function. Therefore, it is physically consistent that the same incident radiation can eject electrons from Q but not from P. The value \(1.91\,\mathrm{eV}\) represents approximately the work function of Q, not the maximum kinetic energy of its emitted electrons. The correct conclusion requires deciding whether emission is allowed before applying the energy-balance equation.

Explanation: \(\textbf{Interpret the graph equation:}\) Einstein's photoelectric equation is \[ K_{\max}=h\nu-\phi \] Since \(\phi=h\nu_0\), it can also be written as \[ K_{\max}=h(\nu-\nu_0) \] Thus, a graph of maximum kinetic energy against incident frequency has slope \(h\). The frequency-axis intercept represents the threshold frequency \(\nu_0\). \(\textbf{Determine Planck's constant from the slope:}\) \[ h=\frac{\Delta K_{\max}}{\Delta\nu} \] Using the two measured points, \[ h=\frac{(2.64-1.32)\times10^{-19}}{(8.0-6.0)\times10^{14}} \] \[ h=\frac{1.32\times10^{-19}}{2.0\times10^{14}} \] \[ h=6.6\times10^{-34}\,\mathrm{J\,s} \] \(\textbf{Calculate the threshold frequency:}\) Use the first measured point in \[ K_{\max}=h(\nu-\nu_0) \] \[ 1.32\times10^{-19}=6.6\times10^{-34}(6.0\times10^{14}-\nu_0) \] Rearranging, \[ 6.0\times10^{14}-\nu_0=\frac{1.32\times10^{-19}}{6.6\times10^{-34}} \] \[ 6.0\times10^{14}-\nu_0=2.0\times10^{14} \] \[ \nu_0=4.0\times10^{14}\,\mathrm{Hz} \] \(\textbf{Convert threshold frequency to wavelength:}\) \[ \lambda_0=\frac{c}{\nu_0} \] \[ \lambda_0=\frac{3.0\times10^8}{4.0\times10^{14}} \] \[ \lambda_0=7.50\times10^{-7}\,\mathrm{m}=750\,\mathrm{nm} \] \(\textbf{Predict the kinetic energy at the new frequency:}\) The new frequency exceeds the threshold frequency, so emission occurs. \[ K_{\max}=h(\nu-\nu_0) \] \[ K_{\max}=6.6\times10^{-34}(5.0\times10^{14}-4.0\times10^{14}) \] \[ K_{\max}=6.6\times10^{-20}\,\mathrm{J} \] \(\textbf{Graph consistency:}\) The frequency \(5.0\times10^{14}\,\mathrm{Hz}\) lies halfway between the threshold frequency and the first measured frequency. Its kinetic energy must consequently be half of \(1.32\times10^{-19}\,\mathrm{J}\), confirming the calculation. The threshold wavelength is a property of the metal, while the calculated kinetic energy corresponds specifically to the new incident frequency. \(\textbf{Final answer:}\) \[ \boxed{\lambda_0=750\,\mathrm{nm},\qquad K_{\max}=6.6\times10^{-20}\,\mathrm{J}} \]

Explanation: \(\textbf{Apply the threshold condition:}\) For each metal, photoelectric emission requires the incident frequency to be at least equal to that metal's threshold frequency. The work function is given by \(\phi=h\nu_0\). Since metal B has a higher threshold frequency than A, B also has the greater work function. \(\textbf{Evaluate Statement I:}\) Radiation X has frequency \(6.0\times10^{14}\,\mathrm{Hz}\). This exceeds the threshold frequency of A but is below the threshold frequency of B. Therefore, radiation X ejects electrons from A but not from B. Statement I is correct. \(\textbf{Evaluate Statement II:}\) Radiation Y has frequency \(8.0\times10^{14}\,\mathrm{Hz}\). This exceeds the threshold frequencies of both metals. Thus, radiation Y can produce photoelectric emission from both A and B. Statement II is correct. \(\textbf{Evaluate Statement III:}\) For a given incident frequency, \[ K_{\max}=h(\nu-\nu_0) \] Metal B has the greater threshold frequency, so a larger fraction of the same incident photon energy is required to overcome its work function. Consequently, photoelectrons from B have smaller maximum kinetic energy than those from A under radiation Y. Statement III is incorrect. \(\textbf{Evaluate Statement IV:}\) The frequency of radiation X remains below the threshold frequency of B. Increasing intensity at fixed frequency increases the incident photon rate but does not increase the energy of an individual photon. Therefore, increasing the intensity of X cannot initiate ordinary photoelectric emission from B. Statement IV is incorrect. \(\textbf{Final inference:}\) The scenario requires two different checks: whether incident photon energy crosses the threshold and, if it does, how much energy remains as photoelectron kinetic energy. The higher-threshold metal is harder to eject electrons from and gives smaller maximum kinetic energy under the same sufficiently energetic radiation. Only Statements I and II are correct.

Explanation: \(\textbf{Identify the independent comparisons:}\) The number of photoelectrons emitted per second depends on the incident photon rate and the fraction of photons that produce emission. The maximum kinetic energy depends on the difference between incident photon energy and the metal's work function. These quantities must be evaluated using different relations. Both photon energies exceed the work function of \(2.0\,\mathrm{eV}\), so photoelectric emission is possible in both cases. \(\textbf{Compare incident photon rates:}\) For a monochromatic source, \[ \dot N=\frac{P}{E_{\mathrm{photon}}} \] For source S1, \[ \dot N_1=\frac{P}{3.0\,\mathrm{eV}} \] For source S2, \[ \dot N_2=\frac{2P}{4.0\,\mathrm{eV}} \] The same energy unit is used in both expressions, so it cancels in the ratio. \[ \frac{\dot N_2}{\dot N_1}=\frac{2P/4.0}{P/3.0} \] \[ \frac{\dot N_2}{\dot N_1}=\frac{2\times3}{4}=\frac{3}{2} \] \(\textbf{Relate photon rate to electron rate:}\) The same fixed fraction of photons produces photoelectrons in both cases. Therefore, the ratio of the photoelectron emission rates is also \(3/2\). The greater radiant power of S2 does not double its photon rate because each S2 photon carries more energy. \(\textbf{Calculate the maximum kinetic energies:}\) Einstein's photoelectric equation gives \[ K_{\max}=E_{\mathrm{photon}}-\phi \] For S1, \[ K_{\max,1}=3.0-2.0=1.0\,\mathrm{eV} \] For S2, \[ K_{\max,2}=4.0-2.0=2.0\,\mathrm{eV} \] Hence, \[ \frac{K_{\max,2}}{K_{\max,1}}=\frac{2.0}{1.0}=2 \] \(\textbf{Physical interpretation:}\) Source S2 produces \(1.5\) times as many photoelectrons per second under the specified equal-emission-fraction condition. Its maximum photoelectron kinetic energy is twice that obtained with S1. The two ratios differ because emission rate depends on incident photon number, whereas maximum kinetic energy depends on the excess energy of each photon above the work function. It would be incorrect to equate the power ratio directly with either the electron-rate ratio or the kinetic-energy ratio. \(\textbf{Final answer:}\) \[ \boxed{\frac{\dot N_{\mathrm{electron},2}}{\dot N_{\mathrm{electron},1}}=\frac{3}{2},\qquad \frac{K_{\max,2}}{K_{\max,1}}=2} \]

Explanation: \(\textbf{Convert all representations into wavelength:}\) The photon energy of electromagnetic radiation is related to wavelength by \[ E=\frac{hc}{\lambda} \] Thus, shorter wavelength corresponds to greater photon energy. Radiation P already has a wavelength of \(400\,\mathrm{nm}\). For radiation Q, the wavenumber is \[ \tilde{\nu}_Q=2.0\times10^4\,\mathrm{cm^{-1}}=2.0\times10^6\,\mathrm{m^{-1}} \] Therefore, \[ \lambda_Q=\frac{1}{\tilde{\nu}_Q}=\frac{1}{2.0\times10^6} \] \[ \lambda_Q=5.0\times10^{-7}\,\mathrm{m}=500\,\mathrm{nm} \] \(\textbf{Determine the wavelength of R:}\) Frequency and time period satisfy \[ \nu_R=\frac{1}{T_R} \] \[ \nu_R=\frac{1}{2.5\times10^{-15}}=4.0\times10^{14}\,\mathrm{Hz} \] Using \(c=\nu\lambda\), \[ \lambda_R=\frac{3.0\times10^8}{4.0\times10^{14}}=750\,\mathrm{nm} \] \(\textbf{Determine the wavelength of S:}\) \[ \lambda_S=\frac{3.0\times10^8}{1.0\times10^{15}}=300\,\mathrm{nm} \] The decreasing order of wavelength is \[ \lambda_R\gt\lambda_Q\gt\lambda_P\gt\lambda_S \] Since photon energy varies inversely with wavelength, the increasing energy order is the reverse of this wavelength order. \(\textbf{Physical interpretation:}\) Radiation S has the shortest wavelength and therefore the greatest photon energy. Radiation R has the longest wavelength and therefore the smallest photon energy. The comparison is valid because all four radiations travel through vacuum with the same propagation speed. A common mistake is to compare the numerical values of frequency, time period and wavenumber directly without converting them into a common physical quantity.

Explanation: \(\textbf{Apply energy quantisation:}\) According to Planck's quantum concept, electromagnetic energy is emitted or absorbed in discrete packets. The energy of one photon is \[ E_{\mathrm{photon}}=\frac{hc}{\lambda} \] A source emitting complete photons must release an integral multiple of its individual photon energy. \(\textbf{Find the photon energies:}\) For source A, \[ E_A=\frac{1240}{620}=2.0\,\mathrm{eV} \] For source B, \[ E_B=\frac{1240}{496}=2.5\,\mathrm{eV} \] \(\textbf{Construct the allowed total energies:}\) Source A can emit positive total energies of \[ 2.0,\ 4.0,\ 6.0,\ 8.0,\ 10.0,\ldots\ \mathrm{eV} \] Source B can emit positive total energies of \[ 2.5,\ 5.0,\ 7.5,\ 10.0,\ldots\ \mathrm{eV} \] The smallest positive energy common to both sequences is \(10.0\,\mathrm{eV}\). \(\textbf{Determine the photon counts:}\) For source A, \[ N_A=\frac{10.0}{2.0}=5 \] For source B, \[ N_B=\frac{10.0}{2.5}=4 \] \(\textbf{Check the quantisation condition:}\) Both photon counts are positive integers, as required. An energy of \(5\,\mathrm{eV}\) can be emitted by B using two photons, but it cannot be emitted by A using complete \(2\,\mathrm{eV}\) photons. Although \(20\,\mathrm{eV}\) is also compatible with both sources, it is not the smallest positive common energy. Equal total radiant energy need not imply equal photon numbers when the photon frequencies differ. \(\textbf{Final answer:}\) \[ \boxed{E_{\min}=10\,\mathrm{eV},\qquad N_A=5,\quad N_B=4} \]