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Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800Å. Calculate threshold frequency (V_(0)) and work function (W_(0)) of metal. |
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Answer» Solution :`"Given "= 6800Å = 6800 xx 10^(-8) cm` `= 6.8 xx10^(-5) cm` `therefore v_(0)=(c)/(lambda)=(3xx10^(10)" cm. SEC"^(-1))/(6.8xx10^(-5)cm)=0.44xx10^(15)s^(-1)` `=4.4xx10^(14)s^(-1)" (or HZ)"` `W_(0)=h lambda_(0)=6.625xx10^(-27)" erg. Sec" xx4.4xx10^(14)S^(-1)`. `29.15xx10^(-13)" erg. For 1 PHOTO ELECTRON."` |
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