. Explain how does (i) photoelectric current and (ii) kinetic energy of the photoelectrons emitted in a photocell vary if the frequency of incident radiation is doubled, but keeping the intensity same? Show the graphical variation in the above two cases.
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(i) photoelectric current depends only on intensity of photons . but here intensity of photons keeping constant { same } .
it means , there is no change in photoelectric current when frequency of incident radiation is doubled.
you can see the graph between photoelectric current and frequency.
(ii) as you know, formula for kinetic energy of emitted photons are given by
K.E = hv - hv₀
here, v is the frequency of incident radiation ,v₀ is the threshold frequency of incident radiation.
now, according to question frequency of incident radiation is doubled then, Kinetic energy also will be greater than double.
[A simple logic ( 6 - 2) = 4 , (12 - 2) = 10 , 2×4+2 >4 × 2 , hence, here it is clear that if frequency is doubled then, Kinetic energy will be greater than double ]
in graph , it is clearly shown kinetic energy and frequency relation.
it means , there is no change in photoelectric current when frequency of incident radiation is doubled.
you can see the graph between photoelectric current and frequency.
(ii) as you know, formula for kinetic energy of emitted photons are given by
K.E = hv - hv₀
here, v is the frequency of incident radiation ,v₀ is the threshold frequency of incident radiation.
now, according to question frequency of incident radiation is doubled then, Kinetic energy also will be greater than double.
[A simple logic ( 6 - 2) = 4 , (12 - 2) = 10 , 2×4+2 >4 × 2 , hence, here it is clear that if frequency is doubled then, Kinetic energy will be greater than double ]
in graph , it is clearly shown kinetic energy and frequency relation.
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