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Monochromatic radiation of wavelength 640.2 nm (1nm=10^(-9)m) fro a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. the source is replaced by an iron source and its 427.2 nm line irradiates the sae photocell. predict the new stopping voltage. |
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Answer» SOLUTION :Here `lamda=640.2nm=6.402xx10^(-7)m,V_(0)=0.54V and lamda.=427.2nm=4.272xx10^(-7)m` Using the relation `hv-phi_(0)=hv-phi_(0)=eV_(0) or (HC)/(lamda)-phi_(0)=eV_(0)`, we have `hc((1)/(lamda.)-(1)/(lamda))=e(V_(0).-V_(0)) or (V_(0).-V_(0))=(hc)/(e)[(1)/(lamda.)-(1)/(lamda)]` `therefore V_(0).=-0.54=(hc)/(e)((1)/(lamda.)-(1)/(lamda))=(6.63xx10^(-34)xx3xx10^(8))/(1.6xx10^(-19))[(1)/(4.272xx10^(-7))-(1)/(6.402xx10^(-7))]=0.97` `implies V_(0).=0.97+0.54=1.51V`. |
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