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When radiation of 400 nm wavelength Is incident on photocell ,emitter produces photoelectron with maximum kinetic energy of 1.68 eV work function of emitter will be………. [hc=1240 eVnm] |
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Answer» 3.09 eV `h_(F)=hf_(0)+(1)/(2)mv_(max)^(2)` `therefore (h_(c))/(lambda)=PHI+(1)/(2)mv_(max)^(2)` `(1240)/(400)=phi+1.68` `therefore phi=3.1-1.68 therefore phi=1.42 eV` |
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