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Ultraviolet light of wavelength 2271 Å fro a 100 W mercury source irradiates a photocell made of molybdenum metal. If the stopping potential is -1.3V, estimate the work function of the metal. How would the photocell respond to a high intensity (~10^(5)Wm^(-2)) red light of wavelength 6328 Å produced bya He-Ne laser ? |
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Answer» Solution :Here `lamda=2271Å=2271xx10^(-10)m=2.271xx10^(-7)m`, and stopping potential `V_(0)=-1.3V` USING Einstein.s photoelectronic equation `hv-phi_(0)=eV_(0)`, we have `phi_(0)=hv-eV_(0)=(hc)/(lamda)-eV_(0)=(6.63xx10^(-34)xx3xx10^(8))/(2.271xx10^(-7))-1.3xx1.6xx10^(-19)J` `=8.758xx10^(-19)-2.08xx10^(-19)J=6.678xx10^(-19)J=(6.678xx10^(-19))/(1.6xx10^(-19))eV=4.17eV` For He-Ne LASER light of wavelength `lamda.=6328Å=6.328xx10^(-7)m,` the enrgy of a PHOTON will be `E.=(hc)/(lamda.)=(6.63xx10^(-34)xx3xx10^(8))/(6.328xx10^(-7))J=(6.63xx10^(-34)xx3xx10^(8))/(6.328xx10^(-7)xx1.6xx10^(-19))eV=1.964eV` As `E.ltphi_(0)`, hence no photoelectric emission will take place IRRESPECTIVE of intenstiy of laser beam. |
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