A metallic surface is illuminated alternatively with light of wavelenghts `3000 Å` and `6000 Å`. It is observed that the maximum speeds of the photoelectrons under these illuminations are in the ratio 3 : 1 . Calculate the work function of the metal and the maximum speed of the photoelectrons in two cases.

A metallic surface is illuminated alternatively with light of waveleng

1.	A metallic surface is illuminated alternatively with light of wavelenghts `3000 Å` and `6000 Å`. It is observed that the maximum speeds of the photoelectrons under these illuminations are in the ratio 3 : 1 . Calculate the work function of the metal and the maximum speed of the photoelectrons in two cases.
Answer» Correct Answer - A::C `E_1 = (12375)/(3000) = 4.125 eV` `E_2 = (12375)/(6000) =2.0625 eV` Maximum speed ratio is 3:1. Therefore, maximum kinetic ratio is 9:1. Now, `9K_(max) = 4.125 - W` `K_(max) = 2.0625 - W` Solving these two equations, we get `W ~~ 1.81 eV. and K_(max) = 0.26 eV` Putting `K_(max) = (1)/(2)mv_(max)^2` we can find `v_(max)`. Here, m is the mass of electron.

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