1.

The frequency of the incident light falling on a photosensitive metal plate is doubled. The kinetic energy of the emitted photoelectron is :

Answer»

unchanged
doubled
less than doubled
more than doubled.

Solution :`E_(k)=hv-phi_(0) rArr E_(k).=2hv-phi_(0)`
`:. (E_(k).)/(E_(k))=(2hv-phi_(0))/(hv-phi_(0))=(2((hv-phi_(0))/(2)))/(hv-phi_(0))s`
Clearly `(E_(k).)/(E_(k)) GT 2`
so `E_(k). gt 2E_(k)`


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