### Related Formula mathrmK.E._max = hnu - phi$\mathrm{K.E.}_{\max} = h\nu - \phi$ ### Core Logic From Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by E_k = hnu - phi$E_k = h\nu - \phi$, where h$h$ is Planck's constant, nu$\nu$ is the incident frequency, and phi$\phi$ is the work function. This represents a straight line equation of the form y = mx + c$y = mx + c$, where y = E_k$y = E_k$ and x = nu$x = \nu$. ### Step 1: Identify the Slope Comparing E_k = hnu - phi$E_k = h\nu - \phi$ with y = mx + c$y = mx + c$, we get the slope: m = tan theta = h$m = \tan \theta = h$ Thus, the slope of the graph gives Planck's constant. ### Pattern Recognition If the y-axis is K.E.$K.E.$, the slope is h$h$. If the y-axis is Stopping Potential V_0$V_0$, the equation is eV_0 = hnu - phi implies V_0 = (h/e)nu - (phi/e)$eV_0 = h\nu - \phi \implies V_0 = (h/e)\nu - (\phi/e)$, and the slope would be h/e$h/e$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Related Formula E = frachclambda$E = \frac{hc}{\lambda}$ lambda_th = frac1240 text eVcdottextnmPhi$\lambda_{th} = \frac{1240 \text{ eV}\cdot\text{nm}}{\Phi}$ ### Core Logic For photoelectric emission to occur, the incident energy must be greater than or equal to the work function (W_e$W_e$ or Phi$\Phi$). The longest wavelength corresponds to the minimum energy required, which is exactly the work function. lambda le frachcW_e$\lambda \le \frac{hc}{W_e}$ ### Step 1: Calculate Wavelength Substitute the given values using the convenient constant hc approx 1240 mathrm~eV cdot nm$hc \approx 1240 \mathrm{~eV \cdot nm}$: lambda le frac1240 mathrm~eV cdot nm3.0 mathrm~eV$\lambda \le \frac{1240 \mathrm{~eV \cdot nm}}{3.0 \mathrm{~eV}}$ lambda le 413.33 mathrm~nm$\lambda \le 413.33 \mathrm{~nm}$ The maximum wavelength (longest wavelength) is: lambda_max approx 414 mathrm~nm$\lambda_{\max} \approx 414 \mathrm{~nm}$ ### Pattern Recognition The standard shortcut E(texteV) = 1240 / lambda(textnm)$E(\text{eV}) = 1240 / \lambda(\text{nm})$ rapidly converts work function to threshold wavelength without dealing with standard SI unit conversions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Related Formula For photoelectric emission to occur: f ge f_0$f \ge f_0$ where f$f$ is incident frequency and f_0$f_0$ is the threshold frequency. ### Core Logic Initially, the frequency is f_1 = 1.5 f_0$f_1 = 1.5 f_0$. Since f_1 > f_0$f_1 > f_0$, emission happens. Then, the frequency is halved: f_2 = frac1.5 f_02 = 0.75 f_0$f_2 = \frac{1.5 f_0}{2} = 0.75 f_0$. ### Step 1: Check Threshold Condition Since f_2 = 0.75 f_0$f_2 = 0.75 f_0$, we see that f_2 < f_0$f_2 < f_0$. The incident light no longer has enough energy per photon to overcome the work function, regardless of how intense the light is. ### Step 2: Conclusion Because the threshold condition fails, emission completely stops. The number of photoelectrons emitted is zero. ### Pattern Recognition Always check the frequency threshold first in photoelectric questions. Intensity adjustments are irrelevant traps if f < f_0$f < f_0$. No emission occurs below threshold frequency. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

### Related Formula frachclambda = phi + K_textmax$\frac{hc}{\lambda} = \phi + K_{\text{max}}$ K_textmax = eV_0$K_{\text{max}} = eV_0$ phi = frachclambda_0$\phi = \frac{hc}{\lambda_0}$ ### Core Logic Using Einstein's Photoelectric equation for the two cases: Case 1 (Wavelength lambda$\lambda$, Stopping Potential 8mathrm\,V$8\mathrm{\,V}$): frachclambda = frachclambda_0 + 8e quad dots dots (texti)$\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + 8e \quad \dots \dots (\text{i})$ Case 2 (Wavelength 3lambda$3\lambda$, Stopping Potential 2mathrm\,V$2\mathrm{\,V}$): frachc3lambda = frachclambda_0 + 2e quad dots dots (textii)$\frac{hc}{3\lambda} = \frac{hc}{\lambda_0} + 2e \quad \dots \dots (\text{ii})$ ### Step 2: Solving the Equations Multiply equation (ii) by 4 to eliminate e$e$: frac4hc3lambda = frac4hclambda_0 + 8e$\frac{4hc}{3\lambda} = \frac{4hc}{\lambda_0} + 8e$ Equating this to equation (i): frachclambda - frachclambda_0 = frac4hc3lambda - frac4hclambda_0$\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{4hc}{3\lambda} - \frac{4hc}{\lambda_0}$ Divide entirely by hc$hc$: frac1lambda - frac1lambda_0 = frac43lambda - frac4lambda_0$\frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{4}{3\lambda} - \frac{4}{\lambda_0}$ frac4lambda_0 - frac1lambda_0 = frac43lambda - frac1lambda$\frac{4}{\lambda_0} - \frac{1}{\lambda_0} = \frac{4}{3\lambda} - \frac{1}{\lambda}$ frac3lambda_0 = frac4 - 33lambda$\frac{3}{\lambda_0} = \frac{4 - 3}{3\lambda}$ frac3lambda_0 = frac13lambda$\frac{3}{\lambda_0} = \frac{1}{3\lambda}$ lambda_0 = 9lambda$\lambda_0 = 9\lambda$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature Of Radiation And Matter