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The electric field associated with a light wave is given by `E=E_(0)sin[(1.57xx10^(7)m^(-1))(ct-x)]`. Find the stopping potential when this light is used in an experiment on a photoelectric effect with the emitter having work function 2.1eV. `h=6.62xx10^(-34)Js`. |
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Answer» Correct Answer - `1.0V` Here, `E=-E_(0)sin[(1.57xx10^(7)m^(-1))(ct-x)]` Comparing it with the equation of harmonic wave in electric field , `E=E_(0)sin(2pi)/lambda(vt-x)` we have , `(2pi)/lambda=1.57xx10^(7) or lambda=(2pi)/(1.57xx10^(7))m`, `V_(0) =(hc)/(lambdae)-(phi_(0))/e` `=(6.63xx10^(-34)xx3xx10^(8)xx1.57xx10^(7))/(2xx3.142xx1.6xx10^(-19))-2.1` `=1.0V` |
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