In an experiment with photoelectric effect, the stopping potential:

### Related Formula h u = phi_0 + K_textmax = phi_0 + eV_0$$h u = \phi_0 + K_{\text{\max}} = \phi_0 + eV_0$$ V_0 = left(frache ight) u - fracphi_0e$$V_0 = \left(\frac{h}{e} ight) u - \frac{\phi_0}{e}$$ ### Core Logic Analyzing each statement: - (A) V_0 text v/s u$V_0 \text{ v/s } u$ is a straight line equation (y = mx + c$y = mx + c$). True. - (B) Slope is frache$\frac{h}{e}$, not fracphi_0h$\frac{\phi_0}{h}$. False. - (C) The slope involves Planck's constant h$h$. True. - (D) To find h$h$ from the slope (m = h/e$m = h/e$), you must multiply the slope by the electronic charge e$e$. Thus, e$e$ is required. False. - (E) The u$ u$-intercept occurs when V_0 = 0 implies u_textth = fracphi_0h$V_0 = 0 \implies u_{\text{th}} = \frac{\phi_0}{h}$. The y$y$-intercept is -phi_0/e$-\phi_0/e$. Therefore, we can find phi_0$\phi_0$ from the intercepts knowing only e$e$ or by looking at the scale parameters carefully, without explicitly knowing h$h$. True. Hence, statements (A), (C), and (E) are correct. ### Pattern Recognition Einstein's photoelectric equation yields a straight-line plot for V_0$V_0$ vs u$ u$ where slope is universally frache$\frac{h}{e}$ for all metals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Related Formula Relativistic total energy equation: E^2 = p^2c^2 + m_0^2c^4$$E^2 = p^2c^2 + m_0^2c^4$$ ### Core Logic We can verify this equation via dimensional analysis: - Dimension of energy, [E] = M^1 L^2 T^-2 implies [E^2] = M^2 L^4 T^-4$[E] = M^1 L^2 T^{-2} \implies [E^2] = M^2 L^4 T^{-4}$ - Dimension of momentum times light speed, [pc] = (M^1 L^1 T^-1) cdot (L^1 T^-1) = M^1 L^2 T^-2 implies [p^2c^2] = M^2 L^4 T^-4$[pc] = (M^1 L^1 T^{-1}) \cdot (L^1 T^{-1}) = M^1 L^2 T^{-2} \implies [p^2c^2] = M^2 L^4 T^{-4}$ - Rest energy square term, [m^2c^4] = M^2 cdot (L^1 T^-1)^4 = M^2 L^4 T^-4$[m^2c^4] = M^2 \cdot (L^1 T^{-1})^4 = M^2 L^4 T^{-4}$ Since all terms share identical dimensions, this relativistic identity is structurally valid. The matching option is (4). ### Pattern Recognition This is Einstein's famous energy-momentum relation of special relativity, fundamental to high-energy modern physics contexts. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Related Formula Power of a light source: P = n E_textphoton = n left(frachclambda ight)$$P = n E_{\text{photon}} = n \left(\frac{hc}{\lambda} ight)$$ where n$n$ is the number of photons emitted per second. ### Core Logic Taking the ratio of power for source 1 and source 2: fracP_1P_2 = fracn_1 left(frachclambda_1 ight)n_2 left(frachclambda_2 ight) = left(fraclambda_2lambda_1 ight) fracn_1n_2$$\frac{P_1}{P_2} = \frac{n_1 \left(\frac{hc}{\lambda_1} ight)}{n_2 \left(\frac{hc}{\lambda_2} ight)} = \left(\frac{\lambda_2}{\lambda_1} ight) \frac{n_1}{n_2}$$ Given data: - fracP_1P_2 = 2$\frac{P_1}{P_2} = 2$ - n_1 = 2 times 10^15\ mathrmphotons/s$n_1 = 2 \times 10^{15}\ \mathrm{photons/s}$ - lambda_1 = 600\ mathrmnm$\lambda_1 = 600\ \mathrm{nm}$ - lambda_2 = 300\ mathrmnm$\lambda_2 = 300\ \mathrm{nm}$ Substitute the values: 2 = left(frac300600 ight) times frac2 times 10^15n_2$$2 = \left(\frac{300}{600} ight) \times \frac{2 \times 10^{15}}{n_2}$$ 2 = frac12 times frac2 times 10^15n_2 implies n_2 = frac10^152 = 5 times 10^14\ mathrmphotons/s$$2 = \frac{1}{2} \times \frac{2 \times 10^{15}}{n_2} \implies n_2 = \frac{10^{15}}{2} = 5 \times 10^{14}\ \mathrm{photons/s}$$ The question asks for the value multiplying 10^14$10^{14}$, which is 5. ### Pattern Recognition Power relies on both the delivery rate and individual photon packet energy. Lower wavelength means more energetic packets, requiring fewer photons to emit the same power. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Related Formula The de Broglie wavelength relation matching a moving particle momentum is given by: lambda = frachp = frachm|vecv| $$\lambda = \frac{h}{p} = \frac{h}{m|\vec{v}|} $$ ### Core Logic Calculate the electric acceleration force acting component on the charge : veca = fracqvecEm = frac(-e)(-E_0hatk)m = fraceE_0mhatk$$\vec{a} = \frac{q\vec{E}}{m} = \frac{(-e)(-E_{0}\hat{k})}{m} = \frac{eE_{0}}{m}\hat{k}$$ Applying kinematics to find velocity at time t$t$ [cite: 127, 718]: vecv(t) = v_0hati + left(fraceE_0tm ight)hatk$$\vec{v}(t) = v_{0}\hat{i} + \left(\frac{eE_{0}t}{m} ight)\hat{k}$$ ### Step 1: Calculating Velocity Magnitude and Final Wavelength Find the magnitude of the updated velocity vector: |vecv| = sqrtv_0^2 + left(fraceE_0tm ight)^2 = v_0sqrt1 + frace^2E_0^2t^2m^2v_0^2 $$|\vec{v}| = \sqrt{v_{0}^{2} + \left(\frac{eE_{0}t}{m} ight)^{2}} = v_{0}\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}} $$ Substitute this into the wavelength equation [cite: 128, 719]: lambda' = frachmv_0sqrt1 + frace^2E_0^2t^2m^2v_0^2$$\lambda' = \frac{h}{mv_{0}\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$$ Since initial wavelength matches lambda_0 = frachmv_0$\lambda_{0} = \frac{h}{mv_{0}}$ , the expression simplifies to : lambda' = fraclambda_0sqrt1 + frace^2E_0^2t^2m^2v_0^2$$\lambda' = \frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$$ ### Pattern Recognition The perpendicular field increases the particle's overall velocity and momentum. Since wavelength is inversely proportional to momentum, it must decrease, which rules out options with a plus sign in the numerator. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Core Logic * **Reflection, Refraction, and Diffraction** can all be fully explained using Huygens' principle and wave theory paths [cite: 54, 47, 66]. * **Compton effect** involves the scattering of an X-ray photon by an electron, demonstrating explicit momentum conversion. This process requires treating light strictly as localized particle packets (photons) and cannot be captured by continuous classical wave formulations. ### Step 1: Conclusion Hence, the Compton effect is the correct answer as it relies entirely on the particle nature of electromagnetic waves. ### Pattern Recognition Phenomena such as the Photoelectric Effect, Compton Scattering, and Blackbody Radiation serve as absolute foundational evidence for the particle/quantum layout of radiation, while Interference, Diffraction, and Polarization confirm wave configurations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter