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The outermost electron of an atom has a binding energy of 2.5 eV. The atom is exposed to light of a high enough frequency to cause exactly one electron to be ejected. The ejected electron is found to have a KE of 2.0 eV. 23. How much energy did photons of the incoming light contain? A) 0.50 eV B) 0.80 eV C) 4.5 eV D) 5.0 eV

User Sugarcrum
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Answer: C) 4.5 eV

Explanation: In order to explain this problem we have to consider the photoelectric effect,

The energy balance in this phenomenum is given by:

h*ν= W+ KE where h*ν is the energy of the incident photon, W is the work function and KE is the kinetic energy of the photoelectrons.

We consider W for outer electrons equal to 2.5 eV , also we consider that the KE is found to be 2.0 eV, then we have:

h*ν=2.5 eV+KE =2.5 eV+2.0 eV0 4.5 eV

User Mhaseeb
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