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Explain giving reasons for the following: (a) Photoelectric current in a photocell increases with the increase in the intensity of the incident radiation. (b) The stopping potential (V_(0)) varies linearly with the frequency (v) of the incident radiation for a given photosensitive surface with the slope remaining the same for different surfaces. (c) Maximum kinetic energy of the photoelectrons is independent of the intensity of incident radiation. |
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Answer» Solution :(a) Photoelectric current in a photocell incrases with increase in number of photoelectrons EJECTED from the CATHODE plate which in turn depends on the number of incident radiation photons. On increasing the intensity of incident radiation the number of incident photons INCREASES and hence photoelectric current increases. (b) As per Einstein.s photoelectric equation, we have `h(v-v_(0))=eV_(0) implies V_(0)=(h)/(e)(v-v_(0))` Thus, `V_(0)` varies linearly with frequency v of the incident radiation for a GIVEN photosensitive surface. the slope of `V_(0)-v` curve is `(h)/(e)` which is a constant and independent of the nature of photosensitive surface. thus, the slope remains the same for different surfaces. (c) As per Einstein.s photoelectric equation, the maximum kinetic energy of ejected phtoelectron is given as `h(v-v_(0))=K_(MAX)` Thus, `K_(max)` depends only on the frequency of incident radiation but is independennt of its intensity. |
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