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

Radiation of frequency 10^(15) Hz is incident on two photosensitive su

1.	Radiation of frequency 10^(15) Hz is incident on two photosensitive surfaces P and Q. there is no photoemission from surface P. photoemission occurs from surface Q but photoelectrons have zero kinetic energy. Explain these observations and find the value of work function for surface Q.
Answer» Solution :Since there is no photoemission from surface P when radiation of frequency `10^(15)` HZ is incident on it, it means that the threshold frequency for surface P is more than `10^(15) `Hz. Thus, `(v_(0))_(P)gt10^(15)`Hz. Again for samme radiation photoemission takes place from surface Q but photoelectrons have zero kinetic energy. so we CONCLUDE that threshold frequency for surface Q is `10^(15)Hz`. thus, `(v_(0))_(Q)=10^(15)` Hz. `therefore`Work FUNCTION for surface `Q=(phi_(0))_(Q)=h(v_(0))_(Q)J=(h(v_(0))_(Q))/(e)EV=(6.63xx10^(-34)xx10^(15))/(1.6xx10^(-19))=4.10eV`

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

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