### Related Formula 1. Magnetic Force on a moving charge: vecF = q(vecv times vecB)$\vec{F} = q(\vec{v} \times \vec{B})$ 2. de-Broglie Wavelength: lambda = frachp = frachm v$\lambda = \frac{h}{p} = \frac{h}{m v}$ where p$p$ is the magnitude of momentum and v$v$ is the speed. ### Core Logic Since the magnetic force vecF$\vec{F}$ is always perpendicular to the velocity vecv$\vec{v}$ of the electron at any instant: W = int vecF cdot dvecr = 0$W = \int \vec{F} \cdot d\vec{r} = 0$ By the work-energy theorem, since work done by the magnetic field is zero, the kinetic energy (and thus the speed v$v$) of the electron remains constant throughout its motion. Since speed v = v_0$v = v_0$ (constant), the magnitude of momentum p = m v$p = m v$ remains constant over time. Therefore, the de-Broglie wavelength remains unchanged: lambda(t) = lambda_0$\lambda(t) = \lambda_0$ ### Pattern Recognition Sees: Charge entering purely magnetic field. Trap: Resolving helical trajectories or cross products mathematically. Do not waste time computing components! Shortcut: A magnetic field can ONLY change the direction of velocity, NEVER the magnitude (speed). Since de-Broglie wavelength depends solely on the magnitude of momentum (p = mv$p = mv$), it must remain constant. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter Class 12 Physics: Moving Charges and Magnetism

### Related Formula K_max = frachclambda - phi$K_{\max} = \frac{hc}{\lambda} - \phi$ p = sqrt2m K_max$p = \sqrt{2m K_{\max}}$ R = fracpeB$R = \frac{p}{eB}$ d = 2R$d = 2R$ ### Core Logic According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectron is: K_max = frachclambda - phi$K_{\max} = \frac{hc}{\lambda} - \phi$ The momentum p$p$ corresponding to this kinetic energy is: p = sqrt2m K_max = sqrt2m left(frachclambda - phiright)$p = \sqrt{2m K_{\max}} = \sqrt{2m \left(\frac{hc}{\lambda} - \phi\right)}$ The electron is emitted normally to the plate and enters a perpendicular constant magnetic field B$B$. It describes a circular arc (semicircle) and hits back the plate at point B. The distance between A and B is the diameter of this circular trajectory: d_AB = 2R = 2 left( fracpeB right) = frac2sqrt2mleft(frachclambda - phiright)eB$d_{AB} = 2R = 2 \left( \frac{p}{eB} \right) = \frac{2\sqrt{2m\left(\frac{hc}{\lambda} - \phi\right)}}{eB}$ To align this with the options, move the factor of 2 inside the square root (2 = sqrt4$2 = \sqrt{4}$): d_AB = fracsqrt4 times 2mleft(frachclambda - phiright)eB = fracsqrt8mleft(frachclambda - phiright)eB$d_{AB} = \frac{\sqrt{4 \times 2m\left(\frac{hc}{\lambda} - \phi\right)}}{eB} = \frac{\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)}}{eB}$ ### Step 1: Final Conclusion The distance between points A and B is $ ### Step 1: Final Conclusion The distance between points A and B is $\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)} / eB. ### Pattern Recognition When a particle is launched perpendicularly from a flat boundary into a perpendicular magnetic field, it describes a semicircle and exits/re-hits the boundary at a distance equal to the diameter $. ### Pattern Recognition When a particle is launched perpendicularly from a flat boundary into a perpendicular magnetic field, it describes a semicircle and exits/re-hits the boundary at a distance equal to the diameter $2R = 2\frac{p}{qB}. Taking coefficients inside square roots converts $. Taking coefficients inside square roots converts $2 \sqrt{2x} to $ to $\sqrt{8x}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Matter and Radiation Class 12 Physics: Moving Charges and Magnetism

### Related Formula lambda = frachp$\lambda = \frac{h}{p}$ p = m v = m (at)$p = m v = m (at)$ a = frace Em = frace sigma^prime2mvarepsilon_0$a = \frac{e E}{m} = \frac{e \sigma^{\prime}}{2m\varepsilon_0}$ where, lambda$\lambda$ = de-Broglie wavelength p$p$ = linear momentum a$a$ = acceleration of the electron in the uniform electric field E$E$ t$t$ = time elapsed since release ### Core Logic Since the electron starts from rest (u = 0$u = 0$), its velocity v$v$ at any time t$t$ is: v = at$v = at$ Thus, the momentum is p = m v = m a t$p = m v = m a t$. Substitute this into the de-Broglie wavelength equation: lambda(t) = frachm a t$\lambda(t) = \frac{h}{m a t}$ Now, compute the rate of change of wavelength with respect to time: fracdlambdadt = fracddt left( frachma t^-1 right) = -frachma t^-2$\frac{d\lambda}{dt} = \frac{d}{dt} \left( \frac{h}{ma} t^{-1} \right) = -\frac{h}{ma} t^{-2}$ This shows that: left| fracdlambdadt right| propto frac1t^2$\left| \frac{d\lambda}{dt} \right| \propto \frac{1}{t^2}$ Comparing this with the given statement (varies inversely as n^textth$n^{\text{th}}$ power of time): n = 2$n = 2$ ### Pattern Recognition Sees: "Uniform electric field" + "de-Broglie wavelength rate of change" → Wavelength lambda propto t^-1$\lambda \propto t^{-1}$. Shortcut: Since lambda propto frac1t$\lambda \propto \frac{1}{t}$, its derivative must scale as fracdlambdadt propto frac1t^2$\frac{d\lambda}{dt} \propto \frac{1}{t^2}$. Thus, n = 2$n = 2$ directly from basic power-rule differentiation! ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Matter and Radiation Class 12 Physics: Electrostatics

### Related Formula K_max = hnu - phi = eV_s$K_{\max} = h\nu - \phi = eV_s$ where, K_max$K_{\max}$ = maximum kinetic energy of photoelectrons V_s$V_s$ = stopping potential e$e$ = fundamental electronic charge ### Core Logic By definition, the stopping potential V_s$V_s$ is the negative potential applied to stop the most energetic photoelectrons from reaching the collector electrode. From Einstein's photoelectric equation: eV_s = K_max implies V_s = fracK_maxe$eV_s = K_{\max} \implies V_s = \frac{K_{\max}}{e}$ Thus, the stopping potential is exactly frac1e$\frac{1}{e}$ times the maximum kinetic energy of the emitted photoelectrons. It does not depend on the intensity of light. ### Pattern Recognition Remember the primary features of the photoelectric effect: - Stopping potential depends linearly on frequency, and inversely on wavelength. - Intensity changes current, but has zero effect on stopping potential. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### 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