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

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Dual Nature Of Radiation And Matter

More Dual Nature of Radiation and Matter Previous-Year Questions — Page 3

Q12 jee_main_2025_24_jan_evening Photoelectric Effect
In photoelectric effect, the stopping potential (V_0) v/s frequency ( u) curve is plotted. (h is the Planck's constant and phi_0 is work function of metal) (A) V_0 v/s u is linear (B) The slope of V_0 v/s u curve = fracphi_0h (C) h constant is related to the slope of V_0 v/s u line (D) The value of electric charge of electron is not required to determine h using the V_0 v/s u curve. (E) The work function can be estimated without knowing the value of h. Choose the correct answer from the options given below :
  • A. (A), (B) and (C) only
  • B. (C) and (D) only
  • C. (A), (C) and (E) only
  • D. (D) and (E) only

Solution

### Related Formula h u = phi_0 + K_textmax = phi_0 + eV_0 V_0 = left(frache ight) u - fracphi_0e ### Core Logic Analyzing each statement: - (A) V_0 text v/s u is a straight line equation (y = mx + c). True. - (B) Slope is frache, not fracphi_0h. False. - (C) The slope involves Planck's constant h. True. - (D) To find h from the slope (m = h/e), you must multiply the slope by the electronic charge e. Thus, e is required. False. - (E) The u-intercept occurs when V_0 = 0 implies u_textth = fracphi_0h. The y-intercept is -phi_0/e. Therefore, we can find phi_0 from the intercepts knowing only e or by looking at the scale parameters carefully, without explicitly knowing h. True. Hence, statements (A), (C), and (E) are correct. ### Pattern Recognition Einstein's photoelectric equation yields a straight-line plot for V_0 vs u where slope is universally frache for all metals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q17 jee_main_2025_24_jan_evening Relativistic Mechanics
The energy E and momentum p of a moving body of mass m are related by some equation. Given that c represents the speed of light, identify the correct equation.
  • A. E^2 = pc^2 + m^2c^4
  • B. E^2 = pc^2 + m^2c^2
  • C. E^2 = p^2c^2 + m^2c^2
  • D. E^2 = p^2c^2 + m^2c^4

Solution

### Related Formula Relativistic total energy equation: E^2 = p^2c^2 + m_0^2c^4 ### Core Logic We can verify this equation via dimensional analysis: - Dimension of energy, [E] = M^1 L^2 T^-2 implies [E^2] = M^2 L^4 T^-4 - Dimension of momentum times light speed, [pc] = (M^1 L^1 T^-1) cdot (L^1 T^-1) = M^1 L^2 T^-2 implies [p^2c^2] = M^2 L^4 T^-4 - Rest energy square term, [m^2c^4] = M^2 cdot (L^1 T^-1)^4 = M^2 L^4 T^-4 Since all terms share identical dimensions, this relativistic identity is structurally valid. The matching option is (4). ### Pattern Recognition This is Einstein's famous energy-momentum relation of special relativity, fundamental to high-energy modern physics contexts. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q23 jee_main_2025_24_jan_evening Photon Theory of Light
The ratio of the power of a light source S_1 to that the light source S_2 is 2. S_1 is emitting 2 times 10^15 photons per second at 600 nm. If the wavelength of the source S_2 is 300 nm, then the number of photons per second emitted by S_2 is ____ times 10^14 .
Numerical Answer. Answer: 5 to 5

Solution

### Related Formula Power of a light source: P = n E_textphoton = n left(frachclambda ight) where n is the number of photons emitted per second. ### Core Logic Taking the ratio of power for source 1 and source 2: fracP_1P_2 = fracn_1 left(frachclambda_1 ight)n_2 left(frachclambda_2 ight) = left(fraclambda_2lambda_1 ight) fracn_1n_2 Given data: - fracP_1P_2 = 2 - n_1 = 2 times 10^15\ mathrmphotons/s - lambda_1 = 600\ mathrmnm - lambda_2 = 300\ mathrmnm Substitute the values: 2 = left(frac300600 ight) times frac2 times 10^15n_2 2 = frac12 times frac2 times 10^15n_2 implies n_2 = frac10^152 = 5 times 10^14\ mathrmphotons/s The question asks for the value multiplying 10^14, which is 5. ### Pattern Recognition Power relies on both the delivery rate and individual photon packet energy. Lower wavelength means more energetic packets, requiring fewer photons to emit the same power. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q17 jee_main_2025_24_jan_morning de Broglie Wavelength of an Electron
An electron of mass 'm' with an initial velocity vecv=v_0hati(v_0>0) enters an electric field vecE=-E_0hatk If the initial de Broglie wavelength is lambda_0, the value after time t would be :-
  • A. fraclambda_0sqrt1+frace^2E_0^2t^2m^2v_0^2
  • B. fraclambda_0sqrt1-frace^2E_0^2t^2m^2v_0^2
  • C. lambda_0
  • D. lambda_0sqrt1+frace^2E_0^2t^2m^2v_0^2

Solution

### Related Formula The de Broglie wavelength relation matching a moving particle momentum is given by: lambda = frachp = frachm|vecv| ### Core Logic Calculate the electric acceleration force acting component on the charge : veca = fracqvecEm = frac(-e)(-E_0hatk)m = fraceE_0mhatk Applying kinematics to find velocity at time t [cite: 127, 718]: vecv(t) = v_0hati + left(fraceE_0tm ight)hatk ### Step 1: Calculating Velocity Magnitude and Final Wavelength Find the magnitude of the updated velocity vector: |vecv| = sqrtv_0^2 + left(fraceE_0tm ight)^2 = v_0sqrt1 + frace^2E_0^2t^2m^2v_0^2 Substitute this into the wavelength equation [cite: 128, 719]: lambda' = frachmv_0sqrt1 + frace^2E_0^2t^2m^2v_0^2 Since initial wavelength matches lambda_0 = frachmv_0 , the expression simplifies to : lambda' = fraclambda_0sqrt1 + frace^2E_0^2t^2m^2v_0^2 ### Pattern Recognition The perpendicular field increases the particle's overall velocity and momentum. Since wavelength is inversely proportional to momentum, it must decrease, which rules out options with a plus sign in the numerator. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q8 jee_main_2025_28_jan_evening Wave Particle Duality
Which of the following phenomena can not be explained by wave theory of light? [cite: 52-53]
  • A. Reflection of light
  • B. Diffraction of light
  • C. Refraction of light
  • D. Compton effect

Solution

### Core Logic * **Reflection, Refraction, and Diffraction** can all be fully explained using Huygens' principle and wave theory paths [cite: 54, 47, 66]. * **Compton effect** involves the scattering of an X-ray photon by an electron, demonstrating explicit momentum conversion. This process requires treating light strictly as localized particle packets (photons) and cannot be captured by continuous classical wave formulations. ### Step 1: Conclusion Hence, the Compton effect is the correct answer as it relies entirely on the particle nature of electromagnetic waves. ### Pattern Recognition Phenomena such as the Photoelectric Effect, Compton Scattering, and Blackbody Radiation serve as absolute foundational evidence for the particle/quantum layout of radiation, while Interference, Diffraction, and Polarization confirm wave configurations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

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