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.

Class 12 Physics in Class 12 3 years ago

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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`.

Posted on 11 Dec 2021, this text provides information on Class 12 related to Physics in Class 12. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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