### Related Formula Einstein's photoelectric equation: eV_s = h u - phi$ eV_s = h u - \phi$ where: * V_s$V_s$ = stopping potential * u$ u$ = frequency of light * phi$\phi$ = work function * Intensity formula: I = fracn h uA cdot t$I = \frac{n h u}{A \cdot t}$ (where n$n$ is rate of photons). ### Core Logic * **Assertion Analysis:** Stopping potential V_s$V_s$ depends strictly linearly on frequency u$ u$ and work function phi$\phi$. It is completely independent of the beam intensity. Therefore, Assertion A is **false**. * **Reason Analysis:** Intensity tracks the flux counts of photons per second. Increasing intensity drives up the quantum count of ejected charges, given u > u_0$ u > u_0$. Thus, Reason R is **true**. ### Pattern Recognition Stopping Potential leftrightarrow$\leftrightarrow$ Frequency/Energy characteristic. Photo-current / Emission Rate leftrightarrow$\leftrightarrow$ Photon Intensity/Flux counts. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Related Formula textK.E = E - W$\text{K.E} = E - W$ [cite: 681] lambda_e = frachsqrt2mtextK.E, quad E = frachclambda_i, quad W = frachclambda_0$\lambda_{e} = \frac{h}{\sqrt{2m\text{K.E}}}, \quad E = \frac{hc}{\lambda_{i}}, \quad W = \frac{hc}{\lambda_{0}}$ [cite: 682] ### Core Logic From the de-Broglie wavelength relationship, squaring both sides yields: [cite: 682] frach^22mlambda_e^2 = textK.E = frachclambda_i - frachclambda_0 = hcleft(frac1lambda_i - frac1lambda_0right)$\frac{h^2}{2m\lambda_{e}^2} = \text{K.E} = \frac{hc}{\lambda_{i}} - \frac{hc}{\lambda_{0}} = hc\left(\frac{1}{\lambda_{i}} - \frac{1}{\lambda_{0}}\right)$ [cite: 682] Simplifying for lambda_e$\lambda_{e}$: [cite: 682] lambda_e^2 = frach^22mhcleft(frac1lambda_i - frac1lambda_0right) = frach2mcleft(frac1lambda_i - frac1lambda_0right)$\lambda_{e}^2 = \frac{h^2}{2mhc\left(\frac{1}{\lambda_{i}} - \frac{1}{\lambda_{0}}\right)} = \frac{h}{2mc\left(\frac{1}{\lambda_{i}} - \frac{1}{\lambda_{0}}\right)}$ lambda_e = sqrtfrach2mcleft(frac1lambda_i - frac1lambda_0right)$\lambda_{e} = \sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_{i}} - \frac{1}{\lambda_{0}}\right)}}$ [cite: 682] ### Pattern Recognition Einstein's photoelectric equation relates kinetic energy linearly to frac1lambda$\frac{1}{\lambda}$ parameters[cite: 681, 682]. Combining this directly into the momentum term sqrt2mK$\sqrt{2mK}$ under the Planck constant yields the standard reciprocal root difference layout[cite: 682]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### 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