### 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

### Related Formula lambda = frachsqrt2mK$\lambda = \frac{h}{\sqrt{2mK}}$ lambda^2 = left(frach^22mright) frac1K$\lambda^2 = \left(\frac{h^2}{2m}\right) \frac{1}{K}$ ### Core Logic Rearranging the de Broglie equation displays a parabolic relationship when evaluating squared attributes or corresponding axes coordinates. Given standard lambda$\lambda$ vs frac1sqrtK$\frac{1}{\sqrt{K}}$ layout tracking, it exhibits an upward facing parabolic behavior matching option (2) layout. ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Related Formula eV_0 = hnu - phi_0$eV_0 = h\nu - \phi_0$ ### Core Logic Assertion (A) is true because applying a negative stopping potential decelerates the emitted photoelectrons and drops the output current down to zero. Reason (R) is true because the stopping potential V_0 = left(fracheright)nu - fracphi_0e$V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$ is linear with frequency nu$\nu$. However, the linearity of V_0$V_0$ vs frequency does not explain the physical mechanism behind why a negative potential stops electron emission. ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter