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The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm ? |
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Answer» Solution :Here work function `phi_(0)=4.2eV` and wavelength of RADIATION `lamda=330nm` Energy of radiation photon `E=(hc)/(lamda)=(6.63xx10^(-34)xx3xx10^(8))/(330xx10^(-9)xx1.6xx10^(-19))eV=3.767eV` As `E lt phi_(0)`, HENCE no PHOTOELECTRIC emission will take place. |
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