Q18.The electric field at a point associated with a light wave is given by E = 200[sin(6 × 1015)t + sin(9 × 1015)t] Vm−1 Given: h = 4. 14 × 10−15 eVs If this light falls on a metal surface having a work function of 2. 50 eV , the maximum kinetic energy of the photoelectrons will be (1) 1. 90 eV (2) 3. 27 eV (3) 3. 60 eV (4) 3. 42 eV
What This Question Tests
This question tests the application of Einstein's photoelectric equation, requiring the calculation of photon energy from given frequencies and then determining the maximum kinetic energy of photoelectrons.
Concepts Tested
Formulas Used
E = hν
K_max = E - Φ
ν = ω / (2π)
📚 NCERT Sections This Tests
11.2 — The Work Function Of Caesium Metal Is 2.14 Ev. When Light Of
Physics Class 12 · Chapter 11
11.2 The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons?
11.8 — Light Of Frequency 7.21 × 1014 Hz Is Incident On A Metal Surface.
Physics Class 12 · Chapter 11
11.8 Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
11.3 — The Photoelectric Cut-Off Voltage In A Certain Experiment Is 1.5 V.
Physics Class 12 · Chapter 11
11.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
📋 Question Details
- Chapter
- Dual Nature of Matter
- Topic
- Photoelectric Effect
- Year
- 2022
- Shift
- 29 Jun Shift 2
- Q Number
- Q18
- Type
- MCQ
- NCERT Ref
- Class 12 Physics Ch 11: Dual Nature of Matter
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