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

Q5 jee_main_2025_02_april_evening de-Broglie Wavelength
An electron with mass m with an initial velocity (t = 0) vecv = v_0hati (v_0 > 0) enters a magnetic field vecB = B_0hatj . If the initial de-Broglie wavelength at t = 0 is lambda_0 then its value after time t would be:
  • A. fraclambda_0sqrt1 - frace^2 B_0^2 t^2m^2
  • B. fraclambda_0sqrt1 + frace^2 B_0^2 t^2m^2
  • C. lambda_0 sqrt1 + frace^2 B_0^2 t^2m^2
  • D. lambda_0

Solution

### Related Formula 1. Magnetic Force on a moving charge: vecF = q(vecv times vecB) 2. de-Broglie Wavelength: lambda = frachp = frachm v where p is the magnitude of momentum and v is the speed. ### Core Logic Since the magnetic force vecF is always perpendicular to the velocity vecv of the electron at any instant: W = int vecF cdot dvecr = 0 By the work-energy theorem, since work done by the magnetic field is zero, the kinetic energy (and thus the speed v) of the electron remains constant throughout its motion. Since speed v = v_0 (constant), the magnitude of momentum p = m v remains constant over time. Therefore, the de-Broglie wavelength remains unchanged: lambda(t) = lambda_0 ### Pattern Recognition Sees: Charge entering purely magnetic field. Trap: Resolving helical trajectories or cross products mathematically. Do not waste time computing components! Shortcut: A magnetic field can ONLY change the direction of velocity, NEVER the magnitude (speed). Since de-Broglie wavelength depends solely on the magnitude of momentum (p = mv), it must remain constant. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter Class 12 Physics: Moving Charges and Magnetism
Q20 jee_main_2025_02_april_morning Photoelectric Effect
A monochromatic light is incident on a metallic plate having work function phi. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is: (Given: The magnitude of charge of an electron is e and mass is m, h is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)
  • A. sqrt2mleft(frachclambda - phiright) / eB
  • B. sqrtmleft(frachclambda - phiright) / eB
  • C. sqrt8mleft(frachclambda - phiright) / eB
  • D. 2 sqrtmleft(frachclambda - phiright) / eB

Solution

### Related Formula K_max = frachclambda - phi p = sqrt2m K_max R = fracpeB d = 2R ### Core Logic According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectron is: K_max = frachclambda - phi The momentum p corresponding to this kinetic energy is: p = sqrt2m K_max = sqrt2m left(frachclambda - phiright) The electron is emitted normally to the plate and enters a perpendicular constant magnetic field B. It describes a circular arc (semicircle) and hits back the plate at point B. The distance between A and B is the diameter of this circular trajectory: d_AB = 2R = 2 left( fracpeB right) = frac2sqrt2mleft(frachclambda - phiright)eB To align this with the options, move the factor of 2 inside the square root (2 = sqrt4): d_AB = fracsqrt4 times 2mleft(frachclambda - phiright)eB = fracsqrt8mleft(frachclambda - phiright)eB ### Step 1: Final Conclusion The distance between points A and B is \sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)} / eB. ### Pattern Recognition When a particle is launched perpendicularly from a flat boundary into a perpendicular magnetic field, it describes a semicircle and exits/re-hits the boundary at a distance equal to the diameter 2R = 2\frac{p}{qB}. Taking coefficients inside square roots converts 2 \sqrt{2x} to \sqrt{8x}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Matter and Radiation Class 12 Physics: Moving Charges and Magnetism
Q21 jee_main_2025_08_april_evening de-Broglie Wavelength
An electron is released from rest near an infinite non-conducting sheet of uniform charge density -sigma^prime. The rate of change of de-Broglie wavelength associated with the electron varies inversely as n^textth power of time. The numerical value of n is
Numerical Answer. Answer: 2 to 2

Solution

### Related Formula lambda = frachp p = m v = m (at) a = frace Em = frace sigma^prime2mvarepsilon_0 where, lambda = de-Broglie wavelength p = linear momentum a = acceleration of the electron in the uniform electric field E t = time elapsed since release ### Core Logic Since the electron starts from rest (u = 0), its velocity v at any time t is: v = at Thus, the momentum is p = m v = m a t. Substitute this into the de-Broglie wavelength equation: lambda(t) = frachm a t Now, compute the rate of change of wavelength with respect to time: fracdlambdadt = fracddt left( frachma t^-1 right) = -frachma t^-2 This shows that: left| fracdlambdadt right| propto frac1t^2 Comparing this with the given statement (varies inversely as n^textth power of time): n = 2 ### Pattern Recognition Sees: "Uniform electric field" + "de-Broglie wavelength rate of change" → Wavelength lambda propto t^-1. Shortcut: Since lambda propto frac1t, its derivative must scale as fracdlambdadt propto frac1t^2. Thus, n = 2 directly from basic power-rule differentiation! ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Matter and Radiation Class 12 Physics: Electrostatics
Q4 jee_main_2025_29_jan_evening Photoelectric Effect and Stopping Potential
In an experiment with photoelectric effect, the stopping potential:
  • A. textincreases with increase in the wavelength of the incident light
  • B. textincreases with increase in the intensity of the incident light
  • C. textis left(frac1mathrmeright) text times the maximum kinetic energy of the emitted photoelectrons
  • D. textdecreases with increase in the intensity of the incident light

Solution

### Related Formula K_max = hnu - phi = eV_s where, K_max = maximum kinetic energy of photoelectrons V_s = stopping potential e = fundamental electronic charge ### Core Logic By definition, the stopping potential V_s is the negative potential applied to stop the most energetic photoelectrons from reaching the collector electrode. From Einstein's photoelectric equation: eV_s = K_max implies V_s = fracK_maxe Thus, the stopping potential is exactly frac1e times the maximum kinetic energy of the emitted photoelectrons. It does not depend on the intensity of light. ### Pattern Recognition Remember the primary features of the photoelectric effect: - Stopping potential depends linearly on frequency, and inversely on wavelength. - Intensity changes current, but has zero effect on stopping potential. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

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