An electron is released from rest near an infinite non-conducting sheet of uniform charge density -sigma^prime$-\sigma^{\prime}$. The rate of change of de-Broglie wavelength associated with the electron varies inversely as n^textth$n^{\text{th}}$ power of time. The numerical value of n$n$ is

### 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

### 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