### 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 The power P$P$ of a light source emitting n$n$ photons per second of wavelength lambda$\lambda$ is given by: P = n cdot frachclambda$P = n \cdot \frac{hc}{\lambda}$ where: * h$h$ is Planck's constant * c$c$ is the speed of light ### Core Logic Since both light sources emit with the same power (P = 200text W$P = 200\text{ W}$), we can relate the number of photons emitted per second for each wavelength: n_1 frachclambda_1 = n_2 frachclambda_2$n_1 \frac{hc}{\lambda_1} = n_2 \frac{hc}{\lambda_2}$ Cancelling out the constant terms h$h$ and c$c$, we get: fracn_1lambda_1 = fracn_2lambda_2 implies fracn_1n_2 = fraclambda_1lambda_2$\frac{n_1}{\lambda_1} = \frac{n_2}{\lambda_2} \implies \frac{n_1}{n_2} = \frac{\lambda_1}{\lambda_2}$ Thus, the ratio of the number of photons emitted is directly proportional to their wavelengths. ### Step 1: Substitute the Values Given values: * lambda_1 = 300text nm$\lambda_1 = 300\text{ nm}$ * lambda_2 = 500text nm$\lambda_2 = 500\text{ nm}$ Substituting these values into the ratio equation: fracn_1n_2 = frac300500 = frac35$\frac{n_1}{n_2} = \frac{300}{500} = \frac{3}{5}$ Thus, the ratio is 3:5$3:5$. ### Pattern Recognition Shortcut: For equal power outputs, the photon emission rate n$n$ is directly proportional to the wavelength lambda$\lambda$. Therefore, ratio of photons n_1 : n_2 = lambda_1 : lambda_2 = 300 : 500 = 3:5$n_1 : n_2 = \lambda_1 : \lambda_2 = 300 : 500 = 3:5$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Core Logic Photoelectric current is directly proportional to the intensity of incident light, which depends on the amount of light energy intercepted. The amount of light energy collected by a lens depends strictly on its aperture diameter. Since both lenses share the exact same diameter, they gather the same total light flux from the source and direct it onto the active photoelectric cell matrix. Thus, the total incident power is invariant. ### Step 1: Conclusion Since the incident energy flux remains constant, the rate of emission of photoelectrons remains identical, meaning the current stays exactly I$I$. ### Pattern Recognition Focal length alters spatial image sizing metrics, but raw aperture dimensions rule total power intercept profiles in optical flux systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter, Ray Optics

### Related Formula For a photon, the energy (which is its kinetic energy) is: E_p = frach clambda_p implies lambda_p = frach cE_p$E_p = \frac{h c}{\lambda_p} \implies \lambda_p = \frac{h c}{E_p}$ For an electron, the de-Broglie wavelength related to its kinetic energy (K_e$K_e$) is: lambda_e = frachm_e v_e = frach v_e2 K_e$\lambda_e = \frac{h}{m_e v_e} = \frac{h v_e}{2 K_e}$ since K_e = frac12 m_e v_e^2 implies m_e v_e = frac2 K_ev_e$K_e = \frac{1}{2} m_e v_e^2 \implies m_e v_e = \frac{2 K_e}{v_e}$. ### Core Logic Given that the velocity of the electron is 25\%$25\%$ of the speed of light: v_e = 0.25c = fracc4$v_e = 0.25c = \frac{c}{4}$ Substituting this into the electron's wavelength equation: lambda_e = frach left(fracc4right)2 K_e = frach c8 K_e$\lambda_e = \frac{h \left(\frac{c}{4}\right)}{2 K_e} = \frac{h c}{8 K_e}$ ### Step 1: Equate Wavelengths We are given that the de-Broglie wavelengths are equal (lambda_p = lambda_e$\lambda_p = \lambda_e$): frach cE_p = frach c8 K_e$\frac{h c}{E_p} = \frac{h c}{8 K_e}$ Cancelling h c$h c$ from both sides: frac1E_p = frac18 K_e implies 8 K_e = E_p$\frac{1}{E_p} = \frac{1}{8 K_e} \implies 8 K_e = E_p$ ### Step 2: Find the Energy Ratio The ratio of the kinetic energy of the electron to that of the photon (E_p$E_p$) is: fracK_eE_p = frac18$\frac{K_e}{E_p} = \frac{1}{8}$ Therefore, the ratio is 1/8$1/8$. ### Pattern Recognition A direct comparison between massive particle kinetic energy (K = fracp^22m = fracp v2$K = \frac{p^2}{2m} = \frac{p v}{2}$) and photon energy (E = p c$E = p c$) when they share the same momentum p$p$ (same wavelength) yields a simple universally applicable rule: fracK_textparticleE_textphoton = fracv2c$\frac{K_{\text{particle}}}{E_{\text{photon}}} = \frac{v}{2c}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter