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The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium, and (b) the wavelengthof the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V. |
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Answer» Solution :a. For the cut-off or THRESHOLD frequency,the energy `h upsilon_(0)` of the incidentradiation must be equal to work function `phi_(0)`, so that `upsilon_(0)=(phi_(0))/(h)=(2.14)/(6.63xx10^(-34))=(2.14xx1.6xx10^(-19))/(6.63xx10^(-34))=5.16xx10^(14)H_(z)` Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected. b. Photocurrent reduces to zero, when maximum kinetic energyof the emitted photoelectrons equals the potential energye `V_(0)` by the retarding potential `V_(0)`. Einstein.s Photoelectric equation is, `eV_(0)=h upsilon-phi_(0)=(hc)/(LAMBDA)-phi_(0) or, lambda=hc//(eV_(0)+phi_(0))` `=((6.63xx10^(-34))xx(3xx10^(8)))/((0.60+2.14))=(19.89xx10^(-26))/((2.74))` `lambda=(19.89xx10^(-26))/(2.74xx1.6xx10^(-19))=454nm` |
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