Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Find the ratio of maximum speeds of emitted electrons.

Explanation

`K_(max) = (1)/(2) mv_(max)^(2) = h upsilon - W_(0)`
`(v_(max)^(2) (1))/(v_(max)^(2) (2)) = ((1-0.5)eV)/((2.5 - 0.5) eV) = (0.5)/(2) = (1)/(4)`
`(v_(max) (1))/(v_(max) (2)) = (1)/(2) = 1 : 2`