In an experiment with photoelectric effect, the stopping potential:

### 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

### 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 de Broglie wavelength relationship to velocity: lambda = frachp = frachmv implies v = frachmlambda$$\lambda = \frac{h}{p} = \frac{h}{mv} \implies v = \frac{h}{m\lambda}$$ ### Core Logic Let mass of proton be m_p$m_p$ and mass of alpha$\alpha$ particle be m_alpha = 4m_p$m_\alpha = 4m_p$. Given wavelengths: lambda_p = lambda$\lambda_p = \lambda$, lambda_alpha = 2lambda$\lambda_\alpha = 2\lambda$. Set up ratios: fracv_pv_alpha = fracm_alpham_p times fraclambda_alphalambda_p$$\frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p} \times \frac{\lambda_\alpha}{\lambda_p}$$ ### Step 1: Substitute Ratios $fracv_pv_alpha = 4 times frac2lambdalambda = 4 times 2 = 8$$$\frac{v_p}{v_\alpha} = 4 \times \frac{2\lambda}{\lambda} = 4 \times 2 = 8$$ Hence, the velocity ratio is $ Hence, the velocity ratio is $8:1. ### Pattern Recognition Remember the standard mass ratio: $. ### Pattern Recognition Remember the standard mass ratio: $m_\alpha \approx 4 m_p$. Inversely proportional components mean velocity amplifies significantly. ### Evaluation Rubric / Model Answer null ### 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