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Let us assume that the current voltage graph for photoelectric tube consists of two straight lines as shown. On such a photoelectric tube, light of wavelength 310 nm is incident. The work function of the emitter plate is 2eV. The tube effectively acts as a solar cell to supply current to an external resistor R. Saturation current is 2A. At the what value of R will be the power dissipated in R maximum |
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Answer» `1Omega` `V=-iR` `V_("stop")=(hc)/(elamda)-(phi)/e=2v` `i=2+V` `I=2-iR` `i(1+R)=2` `i=2/(1+R)` `P=i^(2)R=(4R)/((1+R)^(2))` `(dP)/(DR)=0` `4(1+R)^(2)-4Rxx2(1+R)=0` `4+4R=8R` `R=1Omega` 
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