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Radiation of frequency 1015 Hz is incident on three photosensitive surfaces A, B and C. Following observations are recorded: Surface A: no photoemission occurs - Physics

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प्रश्न

Radiation of frequency 1015 Hz is incident on three photosensitive surfaces A, B and C. Following observations are recorded:

Surface A: no photoemission occurs

Surface B: photoemission occurs but the photoelectrons have zero kinetic energy.

Surface C: photo emission occurs and photoelectrons have some kinetic energy.
Using Einstein’s photo-electric equation, explain the three observations.

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उत्तर

From the observations made (parts A and B) on the basis of Einstein’s photoelectric equation, we can draw the following conclusions:

  1. For surface A, the threshold frequency is more than 1015 HZ, hence no photoemission is possible.
  2. For surface B the threshold frequency is equal to the frequency of given radiation. Thus, photo-emission takes place but the kinetic energy of photoelectrons is zero.
  3. For surface C, the threshold frequency is less than 1015 Hz. So photoemission occurs and photoelectrons have some kinetic energy.
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Einstein’s Photoelectric Equation: Energy Quantum of Radiation
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2022-2023 (March) Sample

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