### 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

### 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