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The K.E. of the most energetic electrons emitted from a metallic surface is doubled when the wavelength, lambda of the incident radiation is reduced from 400 nm to 310 nm. The work function of the metal is |
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Answer» 0.9 eV and `2XX(1)/(2)mv^(2)=(hc)/(lambda)-phi` DIVIDING (ii) by,(i) we get `2=(hc//lambda.-phi)/(hc//lambda-phi)` or `(2hc)/(lambda)-2phi=(hc)/(lambda.)-phi` `or phi=(2hc)/(lambda)-(hc)/(lambda.)` `=(2xx6.63xx10^(-34)xx3xx10^(8))/(4xx10^(-7)xx1.6xx10^(-19))=2.2eV` |
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