In an experiment with photoelectric effect, the stopping potential:

### Core Logic Let mathrmE$\mathrm{E}$ represent the identical energy value shared by both particles: mathrmE_textphoton = fracmathrmhclambda = mathrmE$$\mathrm{E}_{\text{photon}} = \frac{\mathrm{hc}}{\lambda} = \mathrm{E}$$ mathrmE_textproton = fracmathrmp^22mathrmm_mathrmp = mathrmE implies mathrmp = sqrt2mathrmm_mathrmpmathrmE$$\mathrm{E}_{\text{proton}} = \frac{\mathrm{p}^2}{2\mathrm{m}_{\mathrm{p}}} = \mathrm{E} \implies \mathrm{p} = \sqrt{2\mathrm{m}_{\mathrm{p}}\mathrm{E}}$$ Now, expressing the ratio of the proton's de Broglie wavelength to the photon's wavelength: fraclambda_textprotonlambda_textphoton = fracmathrmh/mathrmpmathrmhc/mathrmE = fracmathrmh/sqrt2mathrmm_mathrmpmathrmEmathrmhc/mathrmE$$\frac{\lambda_{\text{proton}}}{\lambda_{\text{photon}}} = \frac{\mathrm{h}/\mathrm{p}}{\mathrm{hc}/\mathrm{E}} = \frac{\mathrm{h}/\sqrt{2\mathrm{m}_{\mathrm{p}}\mathrm{E}}}{\mathrm{hc}/\mathrm{E}}$$ fraclambda_textprotonlambda_textphoton = fracmathrmEmathrmcsqrt2mathrmm_mathrmpmathrmE = frac1mathrmc sqrtfracmathrmE2mathrmm_mathrmp$$\frac{\lambda_{\text{proton}}}{\lambda_{\text{photon}}} = \frac{\mathrm{E}}{\mathrm{c}\sqrt{2\mathrm{m}_{\mathrm{p}}\mathrm{E}}} = \frac{1}{\mathrm{c}} \sqrt{\frac{\mathrm{E}}{2\mathrm{m}_{\mathrm{p}}}}$$ ### Step 1: Final Conclusion The calculated ratio maps to option (3). ### Pattern Recognition Combine the core formulas: lambda_textmatter = fracmathrmhsqrt2mathrmmE$\lambda_{\text{matter}} = \frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}$ and lambda_textlight = fracmathrmhcmathrmE$\lambda_{\text{light}} = \frac{\mathrm{hc}}{mathrm{E}}$. Dividing them smoothly yields the standard non-relativistic scaling ratio. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Related Formula Einstein's Photoelectric Equation: K_textmax = hnu - phi = frachclambda - phi$$K_{\text{max}} = h\nu - \phi = \frac{hc}{\lambda} - \phi$$ For emission to occur, the photon energy must exceed the work function: E_textphoton > phi implies lambda < lambda_textthreshold = frachcphi$$E_{\text{photon}} > \phi \implies \lambda < \lambda_{\text{threshold}} = \frac{hc}{\phi}$$ Useful shortcut: hc approx 1240mathrm~eVcdot nm$hc \approx 1240\mathrm{~eV\cdot nm}$. ### Core Logic Let's find the threshold wavelength lambda_textthreshold$\lambda_{\text{threshold}}$ for a work function of phi = 3mathrm~eV$\phi = 3\mathrm{~eV}$: lambda_textthreshold = frac1240mathrm~eVcdot nm3mathrm~eV approx 413.3mathrm~nm$$\lambda_{\text{threshold}} = \frac{1240\mathrm{~eV\cdot nm}}{3\mathrm{~eV}} \approx 413.3\mathrm{~nm}$$ To cause photoelectric emission, the wavelength of the incident light must be shorter than this threshold: lambda < 413.3mathrm~nm$$\lambda < 413.3\mathrm{~nm}$$ ### Step 1: Comparing Visible Spectrum Colors Let's look at the standard approximate wavelength ranges for visible light: - **Red**: 620 - 750mathrm~nm$620 - 750\mathrm{~nm}$ (Energy approx 1.65 - 2.0mathrm~eV$\approx 1.65 - 2.0\mathrm{~eV}$) - **Yellow**: 570 - 590mathrm~nm$570 - 590\mathrm{~nm}$ (Energy approx 2.1 - 2.2mathrm~eV$\approx 2.1 - 2.2\mathrm{~eV}$) - **Green**: 495 - 570mathrm~nm$495 - 570\mathrm{~nm}$ (Energy approx 2.2 - 2.5mathrm~eV$\approx 2.2 - 2.5\mathrm{~eV}$) - **Blue**: 450 - 495mathrm~nm$450 - 495\mathrm{~nm}$ (At lower end, approaching violet down to 380mathrm~nm$380\mathrm{~nm}$; Energy approx 2.5 - 3.3mathrm~eV$\approx 2.5 - 3.3\mathrm{~eV}$) Only **Blue** light contains wavelengths extending below 413.3mathrm~nm$413.3\mathrm{~nm}$ (high enough photon energy to surpass 3mathrm~eV$3\mathrm{~eV}$). Therefore, blue light is required to cause photoelectric emission from this metal. ### Pattern Recognition Remember standard photon energies of visible colors: Blue/Violet photons have higher energy (typically > 2.8mathrm~eV$> 2.8\mathrm{~eV}$), while Red/Yellow photons have much lower energy (< 2.2mathrm~eV$< 2.2\mathrm{~eV}$). For a higher work function like 3mathrm~eV$3\mathrm{~eV}$, only highly energetic blue/violet light can succeed. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Related Formula Force due to a completely reflecting beam: F = frac2Pc = frac2cfracdEdt$$F = \frac{2P}{c} = \frac{2}{c}\frac{dE}{dt}$$ Change in momentum: Delta p = m v = int F , dt = frac2Ec $$\Delta p = m v = \int F , dt = \frac{2E}{c} $$ Work-Energy Theorem or Conservation of Mechanical Energy for small deflections: gl(2sin^2fractheta2) = fracv^22 $$ gl(2\sin^2\frac{\theta}{2}) = \frac{v^2}{2} $$ For small angles sinfractheta2 approx fractheta2$\sin\frac{\theta}{2} \approx \frac{\theta}{2}$. ### Core Logic Assuming perfect normal reflection from the mirror surface, the pulse imparts a momentum impulse of frac2Ec$\frac{2E}{c}$ to the mass. This provides an initial velocity v$v$ to the mirror. The mirror then swings up to a maximum angle theta$\theta$ where kinetic energy converts entirely to gravitational potential energy.

### Step 1: Calculate Initial Velocity From momentum change: m(v - 0) = frac2Ec implies v = frac2Emc $$ m(v - 0) = \frac{2E}{c} implies v = \frac{2E}{mc}} $$

### Step 2: Relate to Angular Deflection Using conservation of mechanical energy: mgl(1 - costheta) = frac12mv^2$$ mgl(1 - \cos\theta) = \frac{1}{2}mv^2$$ glleft(2sin^2fractheta2 ight) = fracv^22$$glleft(2\sin^2\frac{\theta}{2} ight) = \frac{v^2}{2}$$ Since theta$\theta$ is very small, sinfractheta2 approx fractheta2$\sin\frac{\theta}{2} \approx \frac{\theta}{2}$: glleft(2left(fractheta2 ight)^2 ight) = fracv^22 implies glfractheta^22 = fracv^22 implies gltheta^2 = v^2$$glleft(2\left(\frac{\theta}{2} ight)^2 ight) = \frac{v^2}{2} implies gl\frac{\theta^2}{2} = \frac{v^2}{2} implies gl\theta^2 = v^2$$ ### Step 3: Solve for Theta Substitute v = frac2Emc$v = \frac{2E}{mc}$ into the expression: gltheta^2 = left(frac2Emc ight)^2 = frac4E^2m^2c^2$$gl\theta^2 = \left(\frac{2E}{mc} ight)^2 = \frac{4E^2}{m^2c^2}$$ theta^2 = frac4E^2m^2c^2gl implies theta = frac2Emcsqrtgl $$\theta^2 = \frac{4E^2}{m^2c^2gl} implies \theta = \frac{2E}{mcsqrt{gl}} $$ ### Pattern Recognition This is a standard ballistic pendulum problem where the impulse is delivered by radiation pressure. Perfect reflection means momentum transfer is double the incident momentum (2cdot fracEc$2\cdot \frac{E}{c}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter Class 11 Physics: System of Particles and Rotational Motion

### Related Formula Einstein's photoelectric equation: eV_s = h u - phi$$ eV_s = h u - \phi$$ where: * V_s$V_s$ = stopping potential * u$ u$ = frequency of light * phi$\phi$ = work function * Intensity formula: I = fracn h uA cdot t$I = \frac{n h u}{A \cdot t}$ (where n$n$ is rate of photons). ### Core Logic * **Assertion Analysis:** Stopping potential V_s$V_s$ depends strictly linearly on frequency u$ u$ and work function phi$\phi$. It is completely independent of the beam intensity. Therefore, Assertion A is **false**. * **Reason Analysis:** Intensity tracks the flux counts of photons per second. Increasing intensity drives up the quantum count of ejected charges, given u > u_0$ u > u_0$. Thus, Reason R is **true**. ### Pattern Recognition Stopping Potential leftrightarrow$\leftrightarrow$ Frequency/Energy characteristic. Photo-current / Emission Rate leftrightarrow$\leftrightarrow$ Photon Intensity/Flux counts. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Related Formula textK.E = E - W$$\text{K.E} = E - W$$ [cite: 681] lambda_e = frachsqrt2mtextK.E, quad E = frachclambda_i, quad W = frachclambda_0$$\lambda_{e} = \frac{h}{\sqrt{2m\text{K.E}}}, \quad E = \frac{hc}{\lambda_{i}}, \quad W = \frac{hc}{\lambda_{0}}$$ [cite: 682] ### Core Logic From the de-Broglie wavelength relationship, squaring both sides yields: [cite: 682] frach^22mlambda_e^2 = textK.E = frachclambda_i - frachclambda_0 = hcleft(frac1lambda_i - frac1lambda_0right)$$\frac{h^2}{2m\lambda_{e}^2} = \text{K.E} = \frac{hc}{\lambda_{i}} - \frac{hc}{\lambda_{0}} = hc\left(\frac{1}{\lambda_{i}} - \frac{1}{\lambda_{0}}\right)$$ [cite: 682] Simplifying for lambda_e$\lambda_{e}$: [cite: 682] lambda_e^2 = frach^22mhcleft(frac1lambda_i - frac1lambda_0right) = frach2mcleft(frac1lambda_i - frac1lambda_0right)$$\lambda_{e}^2 = \frac{h^2}{2mhc\left(\frac{1}{\lambda_{i}} - \frac{1}{\lambda_{0}}\right)} = \frac{h}{2mc\left(\frac{1}{\lambda_{i}} - \frac{1}{\lambda_{0}}\right)}$$ lambda_e = sqrtfrach2mcleft(frac1lambda_i - frac1lambda_0right)$$\lambda_{e} = \sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_{i}} - \frac{1}{\lambda_{0}}\right)}}$$ [cite: 682] ### Pattern Recognition Einstein's photoelectric equation relates kinetic energy linearly to frac1lambda$\frac{1}{\lambda}$ parameters[cite: 681, 682]. Combining this directly into the momentum term sqrt2mK$\sqrt{2mK}$ under the Planck constant yields the standard reciprocal root difference layout[cite: 682]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter