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The work function of caesium is 2.14 ev. Find (a) the threshold frequency for caesium, and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V. |
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Answer» Solution :For the cut-off or threshold frequency, the energy h `ν_(0)` of the incident radiation must be equal to work FUNCTION `phi_(0)`, so that `v_(0)=(phi_(0))/(h)=(2.14ev)/(6.63xx10^(-34)Js)` `=(2.14xx1.6xx10^(-19)J)/(6.63xx10^(-3)Js)=5.16xx10^(14)Hz` Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected. (b) Photocurrent reduces to zero, when MAXIMUM kinetic energy of the emitted photoelectrons EQUALS the potential energy `e V_(0)` by the retarding potential `V_(0)` . EINSTEIN’s Photoelectric equation is `eV_(0)=hv-phi_(0)=(hc)/(lambda)-phi_(0)` `or, lambda=hc//(eV_(0)+phi_(0))` `=((6.63xx10^(-34)Js)xx(3xx10^(8)m//s))/((0.60ev+2.14eV))` `=(19.89xx10^(-26)JM)/((2.74eV))` `lambda=(19.89xx10^(-26)Jm)/(2.74xx1.6xx10^(-19)J)=454nm` |
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