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 2

Q18 jee_main_2025_28_jan_morning de Broglie Wavelength
A proton of mass mathrmm_p has same energy as that of a photon of wavelength lambda . If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.
  • A. frac1mathrmcsqrtfrac2mathrmEmathrmm_mathrmp
  • B. frac1c sqrtfracEm_p
  • C. frac1c sqrtfracE2 m_p
  • D. frac12csqrtfracEm_p

Solution

### Core Logic Let mathrmE represent the identical energy value shared by both particles: mathrmE_textphoton = fracmathrmhclambda = mathrmE mathrmE_textproton = fracmathrmp^22mathrmm_mathrmp = mathrmE implies mathrmp = sqrt2mathrmm_mathrmpmathrmE Now, expressing the ratio of the proton's de Broglie wavelength to the photon's wavelength: fraclambda_textprotonlambda_textphoton = fracmathrmh/mathrmpmathrmhc/mathrmE = fracmathrmh/sqrt2mathrmm_mathrmpmathrmEmathrmhc/mathrmE fraclambda_textprotonlambda_textphoton = fracmathrmEmathrmcsqrt2mathrmm_mathrmpmathrmE = frac1mathrmc sqrtfracmathrmE2mathrmm_mathrmp ### Step 1: Final Conclusion The calculated ratio maps to option (3). ### Pattern Recognition Combine the core formulas: lambda_textmatter = fracmathrmhsqrt2mathrmmE and lambda_textlight = fracmathrmhcmathrmE. Dividing them smoothly yields the standard non-relativistic scaling ratio. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q15 jee_main_2025_03_april_morning Photoelectric Effect and Work Function
The work function of a metal is 3mathrm~eV. The color of the visible light that is required to cause emission of photoelectrons is:
  • A. Green
  • B. Blue
  • C. Red
  • D. Yellow

Solution

### Related Formula Einstein's Photoelectric Equation: K_textmax = hnu - phi = frachclambda - phi For emission to occur, the photon energy must exceed the work function: E_textphoton > phi implies lambda < lambda_textthreshold = frachcphi Useful shortcut: hc approx 1240mathrm~eVcdot nm. ### Core Logic Let's find the threshold wavelength lambda_textthreshold for a work function of phi = 3mathrm~eV: lambda_textthreshold = frac1240mathrm~eVcdot nm3mathrm~eV approx 413.3mathrm~nm To cause photoelectric emission, the wavelength of the incident light must be shorter than this threshold: lambda < 413.3mathrm~nm ### Step 1: Comparing Visible Spectrum Colors Let's look at the standard approximate wavelength ranges for visible light: - **Red**: 620 - 750mathrm~nm (Energy approx 1.65 - 2.0mathrm~eV) - **Yellow**: 570 - 590mathrm~nm (Energy approx 2.1 - 2.2mathrm~eV) - **Green**: 495 - 570mathrm~nm (Energy approx 2.2 - 2.5mathrm~eV) - **Blue**: 450 - 495mathrm~nm (At lower end, approaching violet down to 380mathrm~nm; Energy approx 2.5 - 3.3mathrm~eV) Only **Blue** light contains wavelengths extending below 413.3mathrm~nm (high enough photon energy to surpass 3mathrm~eV). Therefore, blue light is required to cause photoelectric emission from this metal. ### Pattern Recognition Remember standard photon energies of visible colors: Blue/Violet photons have higher energy (typically > 2.8mathrm~eV), while Red/Yellow photons have much lower energy (< 2.2mathrm~eV). For a higher work function like 3mathrm~eV, only highly energetic blue/violet light can succeed. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q2 jee_main_2025_04_april_morning Radiation Pressure and Momentum
A small mirror of mass m is suspended by a massless thread of length l. Then the small angle through which the thread will be deflected when a short pulse of laser of energy E falls normal on the mirror (c = speed of light in vacuum and g = acceleration due to gravity)
  • A. theta = frac3E4mcsqrtgl
  • B. theta = fracEmcsqrtgl
  • C. theta = fracE2mcsqrtgl
  • D. theta = frac2Emcsqrtgl

Solution

### Related Formula Force due to a completely reflecting beam: F = frac2Pc = frac2cfracdEdt Change in momentum: Delta p = m v = int F , dt = frac2Ec Work-Energy Theorem or Conservation of Mechanical Energy for small deflections: gl(2sin^2fractheta2) = fracv^22 For small angles sinfractheta2 approx fractheta2. ### Core Logic Assuming perfect normal reflection from the mirror surface, the pulse imparts a momentum impulse of frac2Ec to the mass. This provides an initial velocity v to the mirror. The mirror then swings up to a maximum angle theta where kinetic energy converts entirely to gravitational potential energy.
Deflection of suspended mirror via laser pulse diagram for Q2 - JEE Main 2025 Morning
Deflection of suspended mirror via laser pulse diagram for Q2 - JEE Main 2025 Morning
### Step 1: Calculate Initial Velocity From momentum change: m(v - 0) = frac2Ec implies v = frac2Emc
Deflection of suspended mirror via laser pulse diagram for Q2 - JEE Main 2025 Morning
Deflection of suspended mirror via laser pulse diagram for Q2 - JEE Main 2025 Morning
### Step 2: Relate to Angular Deflection Using conservation of mechanical energy: mgl(1 - costheta) = frac12mv^2 glleft(2sin^2fractheta2 ight) = fracv^22 Since theta is very small, sinfractheta2 approx fractheta2: glleft(2left(fractheta2 ight)^2 ight) = fracv^22 implies glfractheta^22 = fracv^22 implies gltheta^2 = v^2 ### Step 3: Solve for Theta Substitute v = frac2Emc into the expression: gltheta^2 = left(frac2Emc ight)^2 = frac4E^2m^2c^2 theta^2 = frac4E^2m^2c^2gl implies theta = frac2Emcsqrtgl ### Pattern Recognition This is a standard ballistic pendulum problem where the impulse is delivered by radiation pressure. Perfect reflection means momentum transfer is double the incident momentum (2cdot fracEc). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter Class 11 Physics: System of Particles and Rotational Motion
Q10 jee_main_2025_04_april_morning Photoelectric Effect and Intensity
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: In photoelectric effect, on increasing the intensity of incident light the stopping potential increases. Reason R: Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency. In the light of the above statements, choose the correct answer from the options given below
  • A. Both A and R are true but R is NOT the correct explanation of A
  • B. A is false but R is true
  • C. A is true but R is false
  • D. Both A and R are true and R is the correct explanation of A

Solution

### Related Formula Einstein's photoelectric equation: eV_s = h u - phi where: * V_s = stopping potential * u = frequency of light * phi = work function * Intensity formula: I = fracn h uA cdot t (where n is rate of photons). ### Core Logic * **Assertion Analysis:** Stopping potential V_s depends strictly linearly on frequency u and work function phi. It is completely independent of the beam intensity. Therefore, Assertion A is **false**. * **Reason Analysis:** Intensity tracks the flux counts of photons per second. Increasing intensity drives up the quantum count of ejected charges, given u > u_0. Thus, Reason R is **true**. ### Pattern Recognition Stopping Potential leftrightarrow Frequency/Energy characteristic. Photo-current / Emission Rate leftrightarrow Photon Intensity/Flux counts. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q5 jee_main_2025_07_april_evening De Broglie Wavelength of Electron
A photo-emissive substance is illuminated with a radiation of wavelength lambda_i so that it releases electrons with de-Broglie wavelength lambda_c The longest wavelength of radiation that can emit photoelectron is lambda_0. Expression for de-Broglie wavelength is given by: (m: mass of the electron, h: Planck's constant and c: speed of light) [cite: 39, 41]
  • A. lambda_e=sqrtfrach2mc(frac1lambda_i-frac1lambda_o) [cite: 46]
  • B. lambda_c=sqrtfrachlambda_02mc [cite: 47]
  • C. lambda_e=frachsqrt2mc(frac1lambda_i-frac1lambda_o) [cite: 48]
  • D. lambda_c=sqrtfrachlambda_i2mc [cite: 49]

Solution

### Related Formula textK.E = E - W [cite: 681] lambda_e = frachsqrt2mtextK.E, quad E = frachclambda_i, quad W = frachclambda_0 [cite: 682] ### Core Logic From the de-Broglie wavelength relationship, squaring both sides yields: [cite: 682] frach^22mlambda_e^2 = textK.E = frachclambda_i - frachclambda_0 = hcleft(frac1lambda_i - frac1lambda_0right) [cite: 682] Simplifying for lambda_e: [cite: 682] lambda_e^2 = frach^22mhcleft(frac1lambda_i - frac1lambda_0right) = frach2mcleft(frac1lambda_i - frac1lambda_0right) lambda_e = sqrtfrach2mcleft(frac1lambda_i - frac1lambda_0right) [cite: 682] ### Pattern Recognition Einstein's photoelectric equation relates kinetic energy linearly to frac1lambda parameters[cite: 681, 682]. Combining this directly into the momentum term sqrt2mK under the Planck constant yields the standard reciprocal root difference layout[cite: 682]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

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