Q27.The electric field of light wave is given as →E = 10–3 cos( 5×10–72πx −2π × 6 × 1014 t)ˆx NC . This light falls on a metal plate of work function 2 eV . The stopping potential of the photo-electrons is: Given, E (in eV ) = λ(in12375Å) (1) 0.72 V (2) 2.0 V (3) 2.48 V (4) 0.48 V
What This Question Tests
This question combines the properties of an electromagnetic wave (frequency/wavelength from electric field equation) with the photoelectric effect to calculate the stopping potential.
Concepts Tested
Formulas Used
E = hν = hc/λ
E = W₀ + eV_s
V_s = (E - W₀)/e
📚 NCERT Sections This Tests
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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?
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11.6 The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut- off voltage for the photoelectric emission.
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11.9 Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
📋 Question Details
- Chapter
- Dual Nature of Matter
- Topic
- Photoelectric Effect
- Year
- 2019
- Shift
- 09 Apr Shift 1
- Q Number
- Q27
- Type
- MCQ
- NCERT Ref
- Class 12 Physics Ch 11: Dual Nature of Radiation and Matter
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