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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:
- For surface A, the threshold frequency is more than 1015 HZ, hence no photoemission is possible.
- 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.
- For surface C, the threshold frequency is less than 1015 Hz. So photoemission occurs and photoelectrons have some kinetic energy.
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Comment on whether it is consistent with Einstein’s theory:
If c is the velocity of light in free space, the correct statements about photon among the following are:
- The energy of a photon is E = hv.
- The velocity of a photon is c.
- The momentum of a photon, ρ = `(h v)/c`.
- In a photon-electron collision, both total energy and total momentum are conserved.
- Photon possesses positive charge.
Choose the correct answer from the options given below:
