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CBSE Class 12 Physics 2017 Outside Delhi Set 2 Paper

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Question : 6 of 9
Marks: +1, -0
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 (V0)(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.
Solution:  
(a) Variation of photocurrent with intensity of radiation
(b) Stopping potential versus frequency for different materials
(c) Independence of maximum kinetic energy of the emitted photoelectrons
(a) The collision of a photon can cause emission of a photoelectron (above the threshold frequency). As intensity increases, number of photons increases. Hence the current increases.
(b) We have, eVs=h(v−v0)e V_s = h (v - v_0)
∴Vs=he(v)+(−hv0e)\therefore V_s = \frac{h}{e}(v) + \left( -\frac{h v_0}{e} \right)
∴ Graph of VsV_s with vv is a straight line and slope (=he)\left(= \frac{h}{e}\right) is a constant.
(c) Maximum for different surfaces KE=h(v−v0)KE = h (v - v_0)
Hence, it depends on the frequency and not on the intensity of the incident radiation.
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