A monochromatic light is incident on a metallic plate having work function phi$\phi$. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is: (Given: The magnitude of charge of an electron is e$e$ and mass is m$m$, h$h$ is Planck's constant and c$c$ is velocity of light. Take the magnetic field exists throughout the path of electron)

### Related Formula K_textmax = hnu_textmax - phi_0$$K_{\text{max}} = h\nu_{\text{max}} - \phi_0$$ v = fracomega2pi$$v = \frac{\omega}{2\pi}$$ ### Core Logic The light wave consists of two frequencies governed by omega_1$\omega_1$ and omega_2$\omega_2$. omega_1 = 3 times 10^15text rad/s$\omega_1 = 3 \times 10^{15}\text{ rad/s}$ omega_2 = 12 times 10^15text rad/s$\omega_2 = 12 \times 10^{15}\text{ rad/s}$ The maximum kinetic energy of ejected photoelectrons will be determined by the highest frequency photon, which corresponds to omega_2 = 12 times 10^15text rad/s$\omega_2 = 12 \times 10^{15}\text{ rad/s}$. ### Step 1: Calculate Photon Energy Frequency nu_textmax = fracomega_22pi = frac12 times 10^152 times 3.14 approx 1.91 times 10^15text Hz$\nu_{\text{max}} = \frac{\omega_2}{2\pi} = \frac{12 \times 10^{15}}{2 \times 3.14} \approx 1.91 \times 10^{15}\text{ Hz}$. Energy of this photon: E_textphoton = hnu = (6.6 times 10^-34) times (1.91 times 10^15) = 1.26 times 10^-18text J$$E_{\text{photon}} = h\nu = (6.6 \times 10^{-34}) \times (1.91 \times 10^{15}) = 1.26 \times 10^{-18}\text{ J}$$ Convert this energy to eV: E_textmax = frac1.26 times 10^-181.6 times 10^-19 approx 7.87text eV approx 7.9text eV$$E_{\text{max}} = \frac{1.26 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 7.87\text{ eV} \approx 7.9\text{ eV}$$ ### Step 2: Calculate Maximum Kinetic Energy Using Einstein's photoelectric equation: K_textmax = E_textmax - phi_0$$K_{\text{max}} = E_{\text{max}} - \phi_0$$ K_textmax = 7.9 - 2.8 = 5.1text eV$$K_{\text{max}} = 7.9 - 2.8 = 5.1\text{ eV}$$ ### Pattern Recognition When a wave has multiple frequency components (E = E_1sinomega_1 t + E_2sinomega_2 t$E = E_1\sin\omega_1 t + E_2\sin\omega_2 t$), the K_textmax$K_{\text{max}}$ is always strictly determined by the highest frequency (highest energy) component. Ignore the lower frequency terms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### 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 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