Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: In photoelectric effect, on increasing the intensity of incident light the stopping potential increases. Reason R: Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency. In the light of the above statements, choose the correct answer from the options given below

### 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 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