Light is incident on a metallic plate having work function 110 times 10^-20$110 \times 10^{-20}$ J. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is ____ rad/s. (h = 6.63 times 10^-34$h = 6.63 \times 10^{-34}$ J.s)

### 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 lambda = frachp = frachsqrt2mK$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$$ K = qV$K = qV$ ### Core Logic For a charged particle accelerated through a potential difference V$V$, the de Broglie wavelength is: lambda = frachsqrt2mqV$$\lambda = \frac{h}{\sqrt{2mqV}}$$ ### Step 1: Value Substitution lambda = frac6.6 times 10^-34sqrt2 times 6 times 10^-27 times 3 times 10^-19 times 1.21$$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 6 \times 10^{-27} \times 3 \times 10^{-19} \times 1.21}}$$ lambda = frac6.6 times 10^-34sqrt36 times 10^-46 times 1.21$$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{36 \times 10^{-46} \times 1.21}}$$ ### Step 2: Arithmetic Evaluation Evaluate the square root: sqrt36 times sqrt1.21 times sqrt10^-46 = 6 times 1.1 times 10^-23 = 6.6 times 10^-23$$\sqrt{36} \times \sqrt{1.21} \times \sqrt{10^{-46}} = 6 \times 1.1 \times 10^{-23} = 6.6 \times 10^{-23}$$ lambda = frac6.6 times 10^-346.6 times 10^-23$$\lambda = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-23}}$$ lambda = 10^-11 text m$$\lambda = 10^{-11} \text{ m}$$ ### Step 3: Final Conclusion We need to express this in the form alpha times 10^-12 text m$\alpha \times 10^{-12} \text{ m}$: 10^-11 text m = 10 times 10^-12 text m$$10^{-11} \text{ m} = 10 \times 10^{-12} \text{ m}$$ Therefore, alpha = 10$\alpha = 10$. ### Pattern Recognition The expression inside the radical always evaluates cleanly in JEE. 2 times 6 times 3 = 36$2 \times 6 \times 3 = 36$ and 1.21 = (1.1)^2$1.21 = (1.1)^2$ are engineered to perfectly cancel the 6.6$6.6$ in the numerator. ### 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