Dual Nature of Radiation and Matter MCQs with Answers – Part 3 (Class 12 Physics)

Explanation: Row P is correct because at threshold the photon energy \(h\nu_0\) equals the work function. Row Q is correct because using \(c=\nu_0\lambda_0\) gives \( \phi_0=\frac{hc}{\lambda_0} \). Row R is correct because threshold frequency and threshold wavelength refer to the same threshold radiation. Row S is mismatched because \( \lambda_0 \) is inversely proportional to \( \phi_0 \), not directly proportional to it. A metal with a larger work function needs a higher threshold frequency and therefore a smaller threshold wavelength.

Explanation: \( \textbf{Given work function:} \) \( \phi_0=2.8\,eV \). \( \textbf{Planck constant:} \) \(h=4.0\times10^{-15}\,eV\,s\). \( \textbf{Required quantity:} \) Threshold frequency \( \nu_0 \). \( \textbf{Threshold relation:} \) \[ \phi_0=h\nu_0 \] \( \textbf{Rearrangement:} \) \[ \nu_0=\frac{\phi_0}{h} \] \( \textbf{Substitution:} \) \[ \nu_0=\frac{2.8}{4.0\times10^{-15}}\,s^{-1} \] \( \textbf{Coefficient:} \) \[ \frac{2.8}{4.0}=0.70 \] \( \textbf{Power of ten:} \) \[ \frac{1}{10^{-15}}=10^{15} \] \( \textbf{Final simplification:} \) \[ \nu_0=0.70\times10^{15}\,Hz=7.0\times10^{14}\,Hz \] \( \textbf{Final answer:} \) The threshold frequency is \(7.0\times10^{14}\,Hz\).

Explanation: \( \textbf{Threshold frequency:} \) \( \nu_0=5.0\times10^{14}\,Hz \). \( \textbf{Incident frequency:} \) \( \nu=8.0\times10^{14}\,Hz \). \( \textbf{Planck constant:} \) \(h=6.6\times10^{-34}\,J\,s\). \( \textbf{Useful relation:} \) \[ K_{\max}=h(\nu-\nu_0) \] \( \textbf{Why this applies:} \) The work function is \(h\nu_0\), so only the excess frequency above threshold contributes to maximum kinetic energy. \( \textbf{Frequency excess:} \) \[ \nu-\nu_0=3.0\times10^{14}\,Hz \] \( \textbf{Substitution:} \) \[ K_{\max}=(6.6\times10^{-34})(3.0\times10^{14})\,J \] \( \textbf{Coefficient product:} \) \[ 6.6\times3.0=19.8 \] \( \textbf{Power of ten:} \) \[ 10^{-34}\times10^{14}=10^{-20} \] \( \textbf{Final simplification:} \) \[ K_{\max}=19.8\times10^{-20}\,J=1.98\times10^{-19}\,J \] \( \textbf{Final answer:} \) \(K_{\max}=1.98\times10^{-19}\,J\), based on the excess frequency above threshold.

Explanation: The formula compares incident photon energy \( \frac{hc}{\lambda} \) with the work function \( \frac{hc}{\lambda_0} \). For photoemission, the incident photon energy must be at least equal to the work function. This requires \( \lambda\leq\lambda_0 \), because shorter wavelength means higher photon energy. If \( \lambda\gt\lambda_0 \), the expression would give a negative value, which means no photoelectron is emitted rather than a negative kinetic energy. The wavelength condition is the reverse of the frequency condition because frequency and wavelength are inversely related.

Explanation: \( \textbf{Incident wavelength:} \) \( \lambda=400\,nm \). \( \textbf{Threshold wavelength:} \) \( \lambda_0=500\,nm \). \( \textbf{Given constant product:} \) \(hc=1240\,eV\,nm\). \( \textbf{Incident photon energy:} \) \[ E=\frac{hc}{\lambda}=\frac{1240}{400}\,eV=3.10\,eV \] \( \textbf{Work function:} \) \[ \phi_0=\frac{hc}{\lambda_0}=\frac{1240}{500}\,eV=2.48\,eV \] \( \textbf{Maximum kinetic energy:} \) \[ K_{\max}=E-\phi_0 \] \( \textbf{Substitution:} \) \[ K_{\max}=3.10\,eV-2.48\,eV \] \( \textbf{Result:} \) \[ K_{\max}=0.62\,eV \] \( \textbf{Final answer:} \) The maximum kinetic energy is \(0.62\,eV\), and both wavelengths must be used through inverse wavelength.

Explanation: Maximum kinetic energy of a non-relativistic photoelectron is related to its maximum speed by \(K_{\max}=\frac{1}{2}m_ev_{\max}^2\). The right side is in joules when SI units are used, so \(1.6\,eV\) must first be converted into joules for a numerical speed calculation. Since \(1\,eV=1.6\times10^{-19}\,J\), the energy becomes \(1.6(1.6\times10^{-19})\,J\). The expression \(m_ev_{\max}\) represents momentum, not kinetic energy. The relation \(\frac{h}{m_ev_{\max}}\) belongs to de Broglie wavelength, not directly to speed from kinetic energy.

Explanation: \( \textbf{Work function:} \) \( \phi_0=2.0\,eV \). \( \textbf{Stopping potential:} \) \(V_0=1.5\,V\). \( \textbf{Maximum kinetic energy:} \) \[ K_{\max}=eV_0=1.5\,eV \] \( \textbf{Einstein equation:} \) \[ h\nu=\phi_0+K_{\max} \] \( \textbf{Photon energy:} \) \[ h\nu=2.0\,eV+1.5\,eV=3.5\,eV \] \( \textbf{Threshold relation:} \) \[ \phi_0=h\nu_0 \] \( \textbf{Thus:} \) \[ h\nu_0=2.0\,eV \] \( \textbf{Final answer:} \) The incident photon energy is \(3.5\,eV\), while the threshold photon energy is only the work function \(2.0\,eV\).

Explanation: \( \textbf{Frequency-axis intercept:} \) \( \nu_0=6.0\times10^{14}\,Hz \). \( \textbf{Slope of } K_{\max}\textbf{-}\nu \textbf{ graph:} \) \(h=6.6\times10^{-34}\,J\,s\). \( \textbf{Threshold relation:} \) \[ \phi_0=h\nu_0 \] \( \textbf{Why this applies:} \) At the frequency intercept, \(K_{\max}=0\), so the photon energy equals the work function. \( \textbf{Substitution:} \) \[ \phi_0=(6.6\times10^{-34})(6.0\times10^{14})\,J \] \( \textbf{Coefficient product:} \) \[ 6.6\times6.0=39.6 \] \( \textbf{Power of ten:} \) \[ 10^{-34}\times10^{14}=10^{-20} \] \( \textbf{Final simplification:} \) \[ \phi_0=39.6\times10^{-20}\,J=3.96\times10^{-19}\,J \] \( \textbf{Final answer:} \) The work function is \(3.96\times10^{-19}\,J\).

Explanation: Rows P and Q correctly express the work function in terms of threshold frequency and threshold wavelength. Row R correctly follows from \(K_{\max}=h\nu-\phi_0\) and \( \phi_0=h\nu_0 \). Row S has the inverse-wavelength terms in the wrong order. Since \(K_{\max}=E-\phi_0\), the correct wavelength form is \(K_{\max}=hc\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\). This order gives a positive value when the incident wavelength is shorter than the threshold wavelength.

Explanation: At the threshold wavelength, the incident photon energy is exactly equal to the work function. The electron just escapes from the surface, so no excess energy remains for maximum kinetic energy. In equation form, \(K_{\max}=hc\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\). When \( \lambda=\lambda_0 \), the two reciprocal terms are equal and their difference is zero. Threshold emission should therefore be understood as just possible emission, not energetic emission.

Explanation: For a fixed metal, \( \lambda_0=400\,nm \) is the maximum wavelength that can cause emission. Both \(300\,nm\) and \(350\,nm\) are shorter than \(400\,nm\), so both can produce photoemission. However, \(350\,nm\) is longer than \(300\,nm\), so its photon energy is smaller. The excess energy after the work function is therefore smaller at \(350\,nm\). The maximum kinetic energy decreases but does not become zero because the wavelength is still below the threshold wavelength.

Explanation: Kinetic energy of an emitted electron cannot be negative. If substitution in \(K_{\max}=h\nu-\phi_0\) gives a negative value, it means \(h\nu\lt\phi_0\). The incident photon does not have enough energy to liberate an electron from the surface. Therefore, the correct conclusion is no photoelectric emission, not negative motion or negative work function. The formula must be used together with the emission condition \(h\nu\geq\phi_0\).

Explanation: Einstein’s photoelectric equation can be written as \(K_{\max}=h\nu-\phi_0\). This is a straight-line relation between \(K_{\max}\) and \( \nu \). Comparing it with \(y=mx+c\), the coefficient of \( \nu \) is the slope. Therefore, the slope is \(h\), Planck’s constant. The quantity \(\frac{h}{e}\) appears as the slope of a \(V_0\)-\( \nu \) graph, so the vertical-axis quantity must be checked before choosing the slope.

Explanation: The frequency-axis intercept is the point where \(K_{\max}=0\). At this point, the incident photon has just enough energy to remove an electron from the surface. No excess energy remains as kinetic energy, so this is the threshold condition. Hence the intercept gives \( \nu_0 \), the threshold frequency of the metal. Saturation current belongs to a current-potential graph, not to a \(K_{\max}\)-frequency graph.

Explanation: The graph relation is \(K_{\max}=h\nu-\phi_0\). If the straight line is mathematically extended to \( \nu=0 \), the equation gives \(K_{\max}=-\phi_0\). This negative intercept is not a physically measured negative kinetic energy. It shows the energy gap that must first be supplied before emission can begin. The physically meaningful part of the graph for emission starts at \( \nu=\nu_0 \), where \(K_{\max}=0\).

Explanation: \( \textbf{Frequency-axis intercept:} \) \( \nu_0=4.5\times10^{14}\,Hz \). \( \textbf{Planck constant:} \) \(h=6.6\times10^{-34}\,J\,s\). \( \textbf{Required quantity:} \) Work function \( \phi_0 \). \( \textbf{Threshold relation:} \) \[ \phi_0=h\nu_0 \] \( \textbf{Reason:} \) At the frequency-axis intercept, \(K_{\max}=0\), so the photon energy just equals the work function. \( \textbf{Substitution:} \) \[ \phi_0=(6.6\times10^{-34})(4.5\times10^{14})\,J \] \( \textbf{Coefficient product:} \) \[ 6.6\times4.5=29.7 \] \( \textbf{Power of ten:} \) \[ 10^{-34}\times10^{14}=10^{-20} \] \( \textbf{Final simplification:} \) \[ \phi_0=29.7\times10^{-20}\,J=2.97\times10^{-19}\,J \] \( \textbf{Final answer:} \) The work function is \(2.97\times10^{-19}\,J\), obtained from the threshold-frequency intercept.

Explanation: The slope of a \(K_{\max}\)-\( \nu \) graph is \(h\), which is a universal constant. Therefore, different metals should give parallel straight lines when \(K_{\max}\) is plotted against \( \nu \). Their frequency-axis intercepts differ because their threshold frequencies differ. Since \(\phi_0=h\nu_0\), different threshold frequencies mean different work functions. The graph separates a universal slope from a material-dependent intercept.

Explanation: \( \textbf{First graph point:} \) \( \nu_1=5.0\times10^{14}\,Hz \), \(K_1=0\). \( \textbf{Second graph point:} \) \( \nu_2=7.0\times10^{14}\,Hz \), \(K_2=1.32\times10^{-19}\,J\). \( \textbf{Slope of } K_{\max}\textbf{-}\nu \textbf{ graph:} \) \[ h=\frac{\Delta K_{\max}}{\Delta \nu} \] \( \textbf{Change in kinetic energy:} \) \[ \Delta K_{\max}=1.32\times10^{-19}\,J \] \( \textbf{Change in frequency:} \) \[ \Delta \nu=2.0\times10^{14}\,Hz \] \( \textbf{Substitution:} \) \[ h=\frac{1.32\times10^{-19}}{2.0\times10^{14}}\,J\,s \] \( \textbf{Simplification:} \) \[ h=0.66\times10^{-33}\,J\,s=6.6\times10^{-34}\,J\,s \] \( \textbf{Final answer:} \) The graph gives \(h=6.6\times10^{-34}\,J\,s\), with the threshold point acting as one point on the line.

Explanation: The frequency-axis intercept gives the threshold frequency. Metal \(P\) has \( \nu_{0P}=4.0\times10^{14}\,Hz \), while metal \(Q\) has \( \nu_{0Q}=6.0\times10^{14}\,Hz \). For a fixed incident frequency, \(K_{\max}=h(\nu-\nu_0)\). The excess frequency above threshold is larger for metal \(P\) than for metal \(Q\). Therefore, metal \(P\) gives the larger maximum kinetic energy.

Explanation: The equation \(K_{\max}=h\nu-\phi_0\) shows that the energy-axis intercept is \(-\phi_0\). Work function depends on the material and surface condition of the metal. Therefore, different metals generally have different vertical intercepts. The common feature is the slope \(h\), not the intercept. Treating the intercept as universal would incorrectly imply that all metals have the same threshold frequency.