a) =hf=hc/v 5.66 x10-19 J = (6.626 x10-34 Js2)(3.0 x108 m/s2)/v v= 3.51 x10-7 m or 351 nm b) =hf =(6.626 x10-34Js2)(1.24 x1015 Hz) = 8.28 x10-19 J c) first, calculate the work function that 289 nm gives: =hf=hc/v = (6.626 x10-34 Js2)(3.0 x108 m/s2)/ (289 x10-9 m) = 6.88 x10-19 comparing this with the values of given for each metal, you can conclude that electrons will be ejected from cadmium, manganese, and calcium because their respective work functions are lower that 6.88 x10-19 Solution a) =hf=hc/v 5.66 x10-19 J = (6.626 x10-34 Js2)(3.0 x108 m/s2)/v v= 3.51 x10-7 m or 351 nm b) =hf =(6.626 x10-34Js2)(1.24 x1015 Hz) = 8.28 x10-19 J c) first, calculate the work function that 289 nm gives: =hf=hc/v = (6.626 x10-34 Js2)(3.0 x108 m/s2)/ (289 x10-9 m) = 6.88 x10-19 comparing this with the values of given for each metal, you can conclude that electrons will be ejected from cadmium, manganese, and calcium because their respective work functions are lower that 6.88 x10-19.