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Radiation of frequency 10^(15)Hz is inc...

Radiation of frequency `10^(15)Hz` is incident on two photosensitive surface `P` and `Q`. There is no photoemission from surface `P`. Photoemission occurs from `Q` but photoelectrons have zero kinetic energy. Explain these observation and find the value of work function for surface `Q`.

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To solve the problem, we need to analyze the observations regarding the two photosensitive surfaces, P and Q, when radiation of frequency \(10^{15} \text{ Hz}\) is incident on them. ### Step-by-Step Solution: 1. **Understanding Photoelectric Effect**: The photoelectric effect states that when light of sufficient frequency strikes a photosensitive material, it can eject electrons from that material. The energy of the incident photons is given by the equation: \[ E = h \nu ...
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