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Say that I emitted two radiations of same intensity on an anode metal plate( this is regarding the photoelectric effect) . Radiation a has a higher frequency than radiation b , so the kinetic energy gained by electrons from radiation a will be more than b , yet the number of emitted electrons will be the same since both have the same intensity. Now, in which case will the ammeter's reading be more. In other words, will in both cases the current be the same or differ. Note that current is the number of charges over time.

Thank you!

Assuming that each individual light photon has sufficient energy to free the electrons from the surface the current reading will be the same. The only difference will be that in the case of the higher energy [higher frequency] light the emitted electrons will have more energy and, therefore, a higher voltage.


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James J. Kovalcin


I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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