In photoelectric experiment set up, the maximum kinetic energy `(k_("max"))` of emitted photoelectrons was measured for different wavelength `(lambda)` of light used. The graph of `k_("max")` vs `1/(lambda)`   ​ was obtained as shown in first figure. In the same setup, keeping the wavelength of incident light fixed at l, the applied potential difference was varied and the photoelectric current was recorded. The result has been shown in graph in second figure.
(a) Find `lambda` is `Å`
(b) Taking the photo efficiency to be `2%` (i.e. percentage of incident photons which produce photoelectrons) find the power of light incident on the emitter plate in the experiment. [Take `hc = 12400 eV Å`]

Answer

(a) `5536 Å` , (b) `0.089 W`