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If K_(1) and K_(2) are maximum K.E. photoelectron emitted when lights wavelength lambda_(1) and lambda_(2) respectively incident a metallic surface .If lambda_(1)=3lambda_(2),then….. |
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Answer» `K_(2)GT(K_(2))/(3)` `K_(2)=(hc)/(lambda_(2))-phi_(0)` and `(hc)/(lambda_(2))=(K_(2)+phi_(0))`…..(ii) `THEREFORE K_(1)-K_(2)=hc[(1)/(lambda_(1))-(1)/(lambda_(2))]` `=hc[(1)/(3lambda_(2))-(1)/(lambda_(2))]=-(2hc)/(3lambda_(2))` `=-(2)/(3)(K_(2)+phi_(0))[because` From result (1)] `therefore K_(1)=K_(2)-(2)/(3)K_(2)-(2)/(3)phi_(0)=(K_(2))/(3)-(2)/(3)phi_(0)` `therefore K_(1)lt(K_(2))/(3)` |
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