A light wave described by E = 60[sin(3 times 10^15t) + sin(12 times 10^15t)] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) ____ eV. (h = 6.6 times 10^-34text Jcdottexts. and e = 1.6 times 10^-19textC)

Solution & Explanation

### Related Formula K_textmax = hnu_textmax - phi_0 v = fracomega2pi ### Core Logic The light wave consists of two frequencies governed by omega_1 and omega_2. omega_1 = 3 times 10^15text rad/s omega_2 = 12 times 10^15text rad/s The maximum kinetic energy of ejected photoelectrons will be determined by the highest frequency photon, which corresponds to omega_2 = 12 times 10^15text rad/s. ### Step 1: Calculate Photon Energy Frequency nu_textmax = fracomega_22pi = frac12 times 10^152 times 3.14 approx 1.91 times 10^15text Hz. Energy of this photon: E_textphoton = hnu = (6.6 times 10^-34) times (1.91 times 10^15) = 1.26 times 10^-18text J Convert this energy to eV: E_textmax = frac1.26 times 10^-181.6 times 10^-19 approx 7.87text eV approx 7.9text eV ### Step 2: Calculate Maximum Kinetic Energy Using Einstein's photoelectric equation: K_textmax = E_textmax - phi_0 K_textmax = 7.9 - 2.8 = 5.1text eV ### Pattern Recognition When a wave has multiple frequency components (E = E_1sinomega_1 t + E_2sinomega_2 t), the K_textmax is always strictly determined by the highest frequency (highest energy) component. Ignore the lower frequency terms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

Reference Study Guides

Dual Nature Of Radiation And Matter

More Dual Nature of Radiation and Matter Previous-Year Questions — Page 5

Q43 jee_main_2024_29_jan_morning de Broglie Wavelength
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is 25\% of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:
  • A. frac11
  • B. frac18
  • C. frac81
  • D. frac14

Solution

### Related Formula For a photon, the energy (which is its kinetic energy) is: E_p = frach clambda_p implies lambda_p = frach cE_p For an electron, the de-Broglie wavelength related to its kinetic energy (K_e) is: lambda_e = frachm_e v_e = frach v_e2 K_e since K_e = frac12 m_e v_e^2 implies m_e v_e = frac2 K_ev_e. ### Core Logic Given that the velocity of the electron is 25\% of the speed of light: v_e = 0.25c = fracc4 Substituting this into the electron's wavelength equation: lambda_e = frach left(fracc4right)2 K_e = frach c8 K_e ### Step 1: Equate Wavelengths We are given that the de-Broglie wavelengths are equal (lambda_p = lambda_e): frach cE_p = frach c8 K_e Cancelling h c from both sides: frac1E_p = frac18 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) is: fracK_eE_p = frac18 Therefore, the ratio is 1/8. ### Pattern Recognition A direct comparison between massive particle kinetic energy (K = fracp^22m = fracp v2) and photon energy (E = p c) when they share the same momentum p (same wavelength) yields a simple universally applicable rule: fracK_textparticleE_textphoton = fracv2c. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q33 jee_main_2024_30_january_evening Photoelectric Effect Graph
For the photoelectric effect, the maximum kinetic energy left(mathrmE_mathrmkright) of the photoelectrons is plotted against the frequency (nu) of the incident photons as shown in figure. The slope of the graph gives
Photoelectric Effect Graph diagram for Q33 - JEE Main 2024 Evening
A linear graph of maximum kinetic energy E_k versus frequency nu, intersecting the x-axis, with angle theta indicating the slope.
  • A. textRatio of Planck's constant to electric charge
  • B. textWork function of the metal
  • C. textCharge of electron
  • D. textPlanck's constant

Solution

### Related Formula mathrmK.E._max = hnu - phi ### Core Logic From Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by E_k = hnu - phi, where h is Planck's constant, nu is the incident frequency, and phi is the work function. This represents a straight line equation of the form y = mx + c, where y = E_k and x = nu. ### Step 1: Identify the Slope Comparing E_k = hnu - phi with y = mx + c, we get the slope: m = tan theta = h Thus, the slope of the graph gives Planck's constant. ### Pattern Recognition If the y-axis is K.E., the slope is h. If the y-axis is Stopping Potential V_0, the equation is eV_0 = hnu - phi implies V_0 = (h/e)nu - (phi/e), and the slope would be h/e. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q35 jee_main_2024_30_jan_morning Photoelectric Effect and Threshold Wavelength
The work function of a substance is 3.0 mathrm~eV. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:
  • A. 215 mathrm~nm
  • B. 414 mathrm~nm
  • C. 400 mathrm~nm
  • D. 200 mathrm~nm

Solution

### Related Formula E = frachclambda lambda_th = frac1240 text eVcdottextnmPhi ### Core Logic For photoelectric emission to occur, the incident energy must be greater than or equal to the work function (W_e or Phi). The longest wavelength corresponds to the minimum energy required, which is exactly the work function. lambda le frachcW_e ### Step 1: Calculate Wavelength Substitute the given values using the convenient constant hc approx 1240 mathrm~eV cdot nm: lambda le frac1240 mathrm~eV cdot nm3.0 mathrm~eV lambda le 413.33 mathrm~nm The maximum wavelength (longest wavelength) is: lambda_max approx 414 mathrm~nm ### Pattern Recognition The standard shortcut E(texteV) = 1240 / lambda(textnm) 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
Q41 jee_main_2024_31_jan_evening Photoelectric Effect
In a photoelectric effect experiment a light of frequency 1.5 times the threshold frequency is made to fall on the surface of photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photo electrons emitted will be:
  • A. Doubled
  • B. Quadrulated
  • C. Zero
  • D. Halved

Solution

### Related Formula For photoelectric emission to occur: f ge f_0 where f is incident frequency and f_0 is the threshold frequency. ### Core Logic Initially, the frequency is f_1 = 1.5 f_0. Since f_1 > f_0, emission happens. Then, the frequency is halved: f_2 = frac1.5 f_02 = 0.75 f_0. ### Step 1: Check Threshold Condition Since f_2 = 0.75 f_0, we see that 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. No emission occurs below threshold frequency. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q50 jee_main_2024_31_jan_morning Photoelectric Effect
When a metal surface is illuminated by light of wavelength lambda, the stopping potential is 8mathrm\ V. When the same surface is illuminated by light of wavelength 3lambda, stopping potential is 2mathrm\ V. The threshold wavelength for this surface is:
  • A. 5lambda
  • B. 3lambda
  • C. 9lambda
  • D. 4.5lambda

Solution

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

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