Q19.The light of two different frequencies whose photons have energies 3. 8 eV and 1. 4 eV respectively, illuminate a metallic surface whose work function is 0. 6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be : (1) 2 : 1 (2) 4 : 1 (3) 1 : 2 (4) 1 : 4
What This Question Tests
This question applies Einstein's photoelectric equation to calculate the maximum kinetic energy of emitted electrons for two different incident photon energies and then determines the ratio of their maximum speeds.
Concepts Tested
Formulas Used
KE_max = hν - Φ₀
KE_max = (1/2)mv_max²
📚 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
- 24 Jun Shift 2
- Q Number
- Q19
- Type
- MCQ
- NCERT Ref
- Class 12 Physics Ch 11: Dual Nature of Radiation and Matter
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